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fiOBINSON  S  MATHEMATICAL  SERIES.  I 


NEW 

ELEMENTARY  ALGEBRA 


'     i 


OONTAIKINO   THB 


RUDIMENTS  OF  THE  SCIENCE. 


POE 


SCHOOLS     AND    ACADEMIES.  | 


BY 

HORATIO  K  ROBINSON,  LL.  D., 

AUIHO&  OF  A  FULL  COU&SB  OF  KAIHEUATIC8. 


NEW   YUKK 


IVISON,  PHINNEY,  BLAKEMAN  &  CO^ 

CHICAGO  :    S.  C.  GRIGGS  &   CO. 

1866. 


RO  B  IN  so  N'S 


the  most  CoMPLETa,.  jfe£a(/¥EAOTiOAL,  and  most  Scientific  Series  of 
MATHEiiATicfciK  Text-Books  ever  issued  m  this  eouMn, 

^1    ^*6^^£ 

Robinson's  Progressive  Table  Sook,  --•--. 

Robinson's  Progressive  Primary  Arithmetic,-  -  -  o 
Robinson's  j::'rogressive  Intellectual  Arithmetic,  -  «  - 
Robinson's  Rudiments  of  "Written  Arithmetic,  -  -  - 
Robinson's  Progressive  Practical  Arithmetic,  -  -  - 
Robinson's  Key  to  Practical  Arithmetic,  -  -  -  -  - 
Robinson's  Progressive  Higher  Arithmetic,  -  -  -  - 
Robinson's  Key  to  Higher  Arithmetic,    -       -       -       -       - 

Robinson's  Arithmetical  Examples, 

Robinson's  'Nqw  ^Elementary  Algebra,  -  -  .  «  . 
Robinson's  Key  to  Elementary  Algebra,  -       -       -       -       - 

Robinson's  University  Algebra, 

Robinson's  'KeY  to  University  Algebra,  -  -  -  -  - 
Robinson's  I^ew  University  Algebra,  -  -  -  -  - 
Robinson's  Key  to  InTcw  University  Algebra,  -  -  -  - 
Robinson's  IsTew  Grcometry  and  Trigonometry,    -       -       - 

Robinson's  Surveying  and  K"avigation, 

Robinson's  Analyt.  Geometry  and  Conic  Sections, 
Robinson's  Bifferen.  and  Int.  Calculus,  (in  preparatic  i  )- 

Robinson's  Elementary  Astronomy, 

Robinson's  University  Astronomy, 

Robinson's  Mathematical  Operations,        -       -       -       -       „ 

Robinson's  Key  to   Geometry  and  Trigonometry,  Conio 

Sections  and  Analytical  Geometry, 


Entered,  according  to  Act  of  Congress,  in  the  year  1859.  bv 

HOKATIO    N.    ROBINSON,    LL.D., 

In  the  Clerk's  OflSce  of  the  District  Court  of  the  United  States  for  the  Northern 


District  of  New  York. 


PREFACE. 


Within  the  last  twenty  years  Algebra  has  been  steadily 
gainiRg  ground  and  favor  as  an  important  branch  of  education, 
and  it  is  now  taught  and  studied  in  all  the  academies,  semi- 
naries, and  best  public  schools  in  our  country.  While  this  fact 
is  indicative  of  the  onward  progress  of  popular  education,  it 
also  bears  testimony  to  the  value  of  the  science  as  one 
eminently  calculated  to  discipline  the  mind  and  develop  the 
f    reasoning  powers. 

Pupils  now  commence  this  study  at  an  earlier  age  than 
formerly,  and  hence  the  necessity  of  a  work  elementary  in  its 
character,  and  adapted  to  the  comprehension  of  the  youthful 
mind. 

In  the  preparation  of  the  following  treatise  the  author  has 
constantly  kept  in  mind  the  existing  condition  of  school  and 
academic  education,  and  has  adapted  the  work  to  the  most 
approved  modern  methods  of  teaching. 

The  author  believes  this  treatise  to  be  superior  to  other 
elementary  works  upon  the  same  subject  in  the  following  par- 
ticulars :  beauty  of  typography,  the  clear  and  concise  opera^ 
tions  and  a^ialyses  of  the  rules  and  pr^no^p/e5,  the  great 
number  of  examples  and  their  adaptation  to  the  several 
subjectSy  and  the  progressive  character  of  the  work^  so  neces- 
sary to  the  vigorous  development  of  the  intellect. 


IV  PREFACE. 

Particular  attention  is  invited  to  tlie  articles  on  Fractions, 
Simple  and  Higher  Equations,  Powers  and  Roots,  and  the 
analyses  of  the  rules  and  principles,  as  it  is  claimed  that  in 
them  will  be  found  much  that  is  new  and  valuable. 

The  introductory  chapter  is  adapted  to  give  the  pupil  a  cor- 
rect comprehension  of  the  utility  of  symbols,  and  of  the  identity 
and  chain  of  connection  between  Arithmetic  and  Algebra,  in 
which  the  simplicity  of  Mental  Algebra,  and  the  spirit  of  the 
author's  University  Algebra  are  so  blended,  that  the  work  can 
not  fail  to  be  a  most  useful  and  popular  one. 

While  the  author  takes  great  pleasure  in  acknowledging  his 
obligations  to  several  thorough  and  practical  teachers  for  valuable 
hints  and  suggestions,  both  in  theory  and  application,  contributed 
to  this  book,  he  desires  to  make  special  mention  of  the  valuable 
services  rendered  by  J.  C.  Porter,  A.M.,  in  the  preparation  of 
this  work,  whose  acknowledged  ability  as  a  mathematical  scholar, 
and  his  long  and  successful  experience  as  a  teacher,  ought  to 
afford  a  sufficient  guarantee  for  its  practical  character,  and  adap- 
tation to  the  purposes  of  teaching. 

August,  1859. 


CONTENTS. 


SECTIOISr  1. 

DEFINITIONS  AND  NOTATION. 


Th<*  Signs ► page      7     I    Definitions  of  Terms 23 

Definitions  of  Algebraic  Quantities....      9     I    Axioms 24 


ENTIRE  QUANTITIES. 


Addition 25 

Subtraction 33 

Multiplication 40 

To  Square  a  Binomial 47 

To  multiply  the  sum  of  two  quantitiee 

by  their  difference 49 

Division 63 


General  Principles  of  Division 62 

Exact  Division C4 

Reciprocals,  Zero  Powers,  and  Nega- 
tive Exponents 65 

Factoring  Algebraic  Quantities 67 

Greatest  Common  Divisor 72 

Least  Common  Multiple 74 


FRACTIONS. 


Definitions  of  Terms  and  Signs 76 

General  Principles  of  Fractions 77 

Reduction  of  Fractions 78 

Addition 90 


Subtraction 93 

Multiplication 96 

Division 102 

Complex  Fractions...* 105 


SECTION  II. 

SIMPLE    EQUATIONS. 


Definitions 107 

Transformations 108 

Reduction 113 

Proportion,  and  its  relation  to  Equa- 
tions  ".  122 

Problems 124 

Two  Uriknowu  Quantities 139 

1* 


Eliminatioa 140 

Problems  with  two  Unknown  Quan- " 

titie.s 148 

Three  Unknown  Quantities 153 

Problems   with    three  or  more  Un- 
known Quantities 157 

Negative  Results 161 


VI 


CONTENTS. 


SECTION   III. 


POWERS    AND    ROOTS. 


fnvolution 163 

Powers  of  Monomials 164 

Powers  of  Fractions 166 

Powers  of  a  Binomial 168 

Newton's  Binomial  Theorem 171 

French's  Theorem 174 

Evolution 181 

Roots  of  Monomials 181 

Square  Root  of  Polynomials 184 

Square  Root  of  Numbers ]87 

Cube  Root  of  Polynomials 191 


Cube  Root  of  Numbers 195 

Contracted  Method 199 

Reduction  of  Radicals 203 

Addition  of  Radicals 208 

Subtraction  of  Radicals 209 

Multiplication  of  Radicals 210 

Division  of  Radicals 213 

Principles  relating  to  the  application 

of  Involution  and  Evolution 214 

Simple  Equations  containing  Radical 

Quantities 21? 


SECTION  lY. 


QUADRATIC    EQUATIONS- 


Definitions 220 

Pure  Quadratics 220 

problems  producing  Pure  Quadratics.  223 
Affected  Quadratics,  completing  the 

Square 228 

Second    Method    of    completing    the 

Square 232 

Higher   Equations  in  the  Quadratic 

Form 236 

Polynomials    under    the    Quadratic 

Form 238 


Formation  of   Quadratic  Equations, 

First  Method 239 

Second  Method 241 

Factoring  Trinomials 243 

Rfc)al  and  Imaginary  Roots 244 

Problems  producing  Quadratic  Equa- 
tions   245 

Quadratic   Equations  containing  two 

Unknown  Quantities 247 

Problems  producing  Quadratics  with 
two  Unknown  Quantities 2;'''2 


SECTION  Y. 

PROGRESSIONS  AND  PROPORTIONS. 


Arithmetical  Progression 255 

General  Applications 268 

Problems 260 

SpeciUl  Applications 261 

Geometrical  Progression 205 


In-lnite  Series 268 

Geometrical  Means 270 

Applications < 272 

Problems 272 

Special  Applications .t...  27^ 


V 


PROPORTION. 


Definitions „ 277 

General  Principlefl,  in  Twelve  Propositions 279 

Problems 286 

Approximate  Roots  op  ITigher  Deqeees 289 

Miscellaneous  Examples 297, 


\ 


ELEMENTAEY  ALGEBRA. 


SECTION  1. 
DEFINITIONS  AND  NOTATION. 

1.  Guantity  is  anything  that  can  be  measured  or  compared ; 
as  distance,  space,  motion,  time. 

3.  Mathematics  is  the  science  which  treats  of  the  relations 
of  quantities. 

3.  Algebra  is  that  branch  of  mathematics  in  which  the  ope- 
rations are  indicated  by  signs  or  symbols,  and  the  quantities 
are  represented  by  letters.     It  is  universal  arithmetic. 

THE   SIGNS. 

4.  Addition  is  denoted  by  the  perpendicular  cross,  +,  called 
plus  ;  thus,  in  4  +  2  -f  9,  the  sign  indicates  that  4,  2,  and  9 
are  to  be  added. 

^.  Subtraction  is  denoted  by  the  horizontal  line,  — ,  called 
minus  ;  thus,  in  10  —  7,  the  sign  indicates  that  7,  the  number 
after  it,  is  to  be  subtracted  from  10,  the  number  before  it. 

5.  Multiplication  is  denoted  by  the  oblique  cross,  X  ;  tbus^ 
in  5  X  4,  the  sign  indicates  that  5  and  4  are  to  be  multiplied 
together. 

'^.  Division  is  denoted  by  a  horizontal  line,  with  a  point 
above  and  one  below,  —- ;  thus,  in  18  -f-  6,  the  sign  indicates 

Pefine  quantity.  Mathematics.  Algebra.  Explain  the  sign  of  Addi- 
tion,    Of  Subtraction.     Multiplication.     Division 

(7) 


8  ALGEBRAIC  QUANTITIES. 

that  18,  the  number  before  it,  is  to  be  divided  by  6,  the  number 

after  it.     The  horizontal  line  without  the  points  becomes  the 

sign  of  division  when  the  dividend  is  written  above  and  the 

18 
divisor  below  it;    thus,  -^  indicates  division,  the  same  as 

18  -f-  6. 

8*  Equality  is  denoted  by  two  horizontal  parallels,  =,  which 
represent  the  words,  equal  to ;  thus,  in  4  +  8  =  7  +  5;  the 
parallels  indicate  the  equality  of  the  two  sets  of  numbers  com- 
pared ;  and  the  expression  is  read,  4  plus  8  are  equal  to  7  plus  5. 

0,  Inequality  is  denoted  by  the  angle,  ^,  the  opening 
always  being  toward  the  larger  number  or  quantity ;  thus,  in 
12  -f  7  >  14,  the  sign,  ^,  indicates  that  the  sum  of  12  and  7  is 
greater  than  14,  and  the  whole  expression  is  read,  12  plus  7  is 
greater  than  14.  The  expression  6  <^  4  -f  7  is  read,  6  is  less 
than  4  plus  7. 

10.  A  parenthesis,  (  ),  denotes  that  the  several  numbers 
or  quantities  included  within  it  are  to  be  considered  together, 
and  subjected  to  the  same  operation  ;  thus,  (10  -f  4)  x  3  indi- 
cates that  both  10  and  4,  or  their  sum,  is  to  be  multiplied  by 
8 ;  (10  —  4)  X  3  indicates  that  the  difference  of  10  and  4  is 
to  be  multiplied  by  3. 

11.  A  horizontal  vinculum, ,  placed  over  the  num- 
bers or  quantities,  is  frequently  used  instead  of  the  parenthe- 
Bis ;  thus,  4+  2-^3  X  7  is  equivalent  to  (4  +  2  +  3)  x  7. 

12.  The  radical  sign,  V ,  indicates  that  the  root  of  the 
quantity  placed  under  it  is  to  be  taken. 

EXAMPLES. 

1.  4-1-7  —  6  =^  how  many  ?  Arts. 

2,  36  —  5  —  20  -f  1  =  how  many  ?  Ans, 

3.  — ^p—  =  how  many  ?  Ans.  3, 


5. 

Ans,  12. 


Explain  the  sign  of  Equality.     Inequality.     Explain  the  use  of  the 
Parenthesis.     Of  the  Vinculum.     The  Radical  Sign. 


DEFIi;iTIONS  AND  NOTATION.  Q 

A^  Q      I     2 

4.  ~ — :  X  7  =  how  many  ?  Ans,  98, 

72  —  2      5  X  IS       ,  „  .IP 

6.  — ^ — I irr —  =  now  many  r  -4ns.  lb. 

.    86— (8  x2)       7  +  21      ,  ,  ,        ^ 

I).  ^ — —  =  now  many  r  Ans.  2. 

7.  (4  +  6)  X  12  =  how  many  ?  Jws.  120. 

8.  Show  that  (75  —  25)  -^  25  =  2. 

o    Qi.       .1      100  +  30  —  70       ^ 

9.  Show  that -^ =  1. 

oO 

10.  Show  that  (^?i:^?/—  6)  x  5  =  25. 

11.  Show  that  (17  +  4  — 11)  x  6  >  50. 

12.  Show  that  (?^I^-^  ^  6)  +  18  <i 

13.  Show  that  ^^jL^i^J  >  (16  — 12  +  10)  X  5  —  67, 


1674  —  1569   ^/1765  — 1653^ 

7' 


14.   Show  that  ;:; <f --—■ — j  X  7. 


ALGEBRAIC   QUANTITIES. 

15.  The  quantities  considered  in  Algebra  are  of  two  kinds, 
known  and  unknoivn. 

14.  Known  Quantities  are  those  whose  values  are  given ; 
they  are  represented  by  the  first  letters  of  the  alphabet,  as 
a,  6,  c,  d. 

lo.  Unknown  Quantities  are  those  whose  values  are  to  be 
determined ;  they  are  represented  by  the  final  letters  of  the 
alphabet,  as  tt,  x^  y,  z. 

16,  Literal  Quantities  are  those  numbers  or  values  which 
a.re  expressed  by  letters. 

\7*  The  multiplication  of  literal  quantities  is  expressed  by 
^mply  writing  the  factors  together,  without  the  sign  x  ;  thus, 
2  times  x  is  written  2x ;  3  times  x  is  written  ^x ;  a  times  x  is 
written  ax ;  and  a  times  h  times  y  is  written  aby ;  and  so  on. 

Define  known  quantities.  Unknown  quantities.  Literal  quan-'tiea 
How  is  the  multiplication  of  literal  quantities  expressed  ? 


10  ALGEBRAIC  QUANTITIES. 

18.  A  Coefiicieiit  is  a  number  or  quantity  prefixed  to  an 
other  quantity,  to  denote  how  many  times  the  latter  is  taken ; 
thus,  in  ox,  3  is  the  coefficient  of  ^,  and  indicates  that  x  is 
taken  3  times ;  in  ax,  a  is  the  coefficient  of  x,  and  indicates 
that  X  is  taken  a  times  ;  in  Zax,  8  may  1)e  regarded  as  the  co- 
efficient of  ax,  or  3a  as  the  coefficient  of  x  ;  in  4  (a  +  a;),  4  ig 
the  coefficient  of  (a  +  x).  When  no  coefficient  is  written,  the 
iiiit  1  is  understood. 

m.  An  Exponent  is  a  number  written  above,  and  to  the 

right  of  a  quantity,  to  denote  how  many  times  the  quantity  is 

used  as  a  factor ;  thus,  the  repetition  of  the  same  factor,  as 

XXX,  may  be  written  x^,  in  which  3  is  the  exponent  of  x,  and 

indicates  that  x  is  used  3  times  as  a  factor. 

Note.  —  To  enable  tlie  pnpil  to  discriminate  between  a  coefficient  and 
an  qxponont,  the  two  may  be  contrasted,  thus  :  Zx  indicates  that  x  is 
multiplied  by  3 ;  x^  indicates  that  x  is  multiplied  by  itself  till  it  is  used 
3  times  as  a  factor.  Again,  Sz  is  an  abbreviation  oi  x-^-  x-\-  x;  x^  is 
an  abbreviation  of  a;  x  ^  X  2:,  or  of  a:  a:  a:. 

S#,  A  Power  is  the  product  obtained  by  repeating  a  quan- 
tity several  times  as  a  factor,  and  is  indicated  hj  an  exponent ;  x. 
thus,  a-  is  the  second  power  of  a,  or  a  a ;  o-'  tlie  third  power, 
or  a  a  a  ;  a*  the  fourth  power,  or  a  a  aa.     Wlien  no  exponent 
is  written,  1  is  understood. 

Si.  A  Root  is  a  factor  repeated  to  form  a  power ;  thus,  in 
m^,  m  is  the  root  which  is  repeated  to  form  the  power  iiv*. 

S^.  An  Equation  is  an  expression  of  equality  between  two 
quantities  ;  thus,  as'  =4  ;  5^7  =  60  ;  3^  =  a  +  h,  are  equations. 

23.  The  First  Member  of  an  equation  is  the  part  on  the 
left  of  the  sign  =. 

24.  The  Second  Member  of  an  equation  is  the  part  on  the 
right  of  the  sign  =. 

To  exercise  the  pupil  in  Algebraic  IN'otation,  and  show  him 
the  use  of  symbols  and  algebraic  forms,  we  introduce  here, 
under  strict  classification,  the  following 

Defi'c  a  Coefficient.  An  Exponent  A  Power.  A  Root.  An  Equation, 
and  '  ^  members- 


DEFINITIONS  AND  NOTATION. 


11 


EXAMPLES   FOR   PHACTICE. 


g5,  1.  A  man  bought  a  coat  and  a  hat  for  12  dollars,  and 
the  coat  cost  twice  as  much  as  the  hat ;  what  was  the  price 
of  each? 


Let 


OPERATION. 

:  price  of  the  hat ; 
:  price  of  the  coat. 


Analysis.  The  hat  cost  a  cer- 
tain number  of  dollars,  which, 
when  known,  will  determine  the 
price  of  the  coat.  For  brevity 
and  distinctness,  we  represent  the 
price  of  the  hat  by  the  letter  x ; 
and  as  the  coat  cost  2  times  as 
much,  2a:  will  represent  the  price 
of  the  coat.  Since  2a:  added  to  x  are  3a:,  the  coat  and  hat  toge- 
ther cost  3a:.  But  by  the  conditions  of  the  example,  the  coat  and  hat 
together  cost  12  dollars;  hence,  3a:  are  equal  to  12  dollars;  and  a:, 
the  price  of  the  hat,  must  be  \  of  12,  dollars,  or  4  dollars;  and  2a:, 
the  price  of  the  coat,  must  be  2  times  4  dollars,  or  8  dollars. 


6X 

X  ■ 

2x-. 


12  dollars. 
4  dollars,  hat ; 
8  dollars,  coat. 


2.  A  man  bought  a  saddle  and  bridle  for  45  dollars ;  the 
saddle  cost  4  times  as  much  as  the  bridle ;  what  was  the  price 
of  each  ? 


OPERATION. 

X  =  price  of  the  bridle ; 
4^  =  price  of  the  saddle. 
5d7  =  45  dollars. 

^  ==    9  dollars,  bridle  ; 
Ax  —  ZQ  dollars,  saddle. 


Analysis.   We  represent  the 
Let    X  =  price  of  the  bridle  ;  P^'i^e  of  the  bridle  by  the  letter 

a: ;  and  as  the  saddle  cost  4  times 
as  much  as  the  bridle,  4a:  will 
represent  the  price  of  the  sad- 
dle. Since  4x  added  to  x 
are  5x,  both  saddle  and  bridle 
together  cost  5x.  But  by  the 
conditions  of  the  example,  the  saddle  and  bridle  together  cost  45 
dollars ;  hence,  5a;  are  equal  to  45  dollars ;  and  x,  the  price  of  the 
bridle,  must  be  J  of  45  dollars,  or  9  dollars  ;  and  4a;,  the  price  of  the 
Eaddle,  must  be  4  times  9  dollars,  or  36  dollars. 

-     o    The  greater  of  two  numbers  is  8  times  the  less,  and  their 
sum  is  108  ;  what  are  the  two  numbers  ? 


22  ALQEBEAIC  QUANTITIES 

orERATlON.  Analysis.      We  represent  the  less 

Let    X  =  the  less  •  number  by  x ;  and  as  the  greater  is  8 

8:r  =  the  greater.         *'">««  ^^l  1«'«;,?'.  ^"^  ^' ''/irf^f 

by  Sx.     By  addition,  we  find  that  the 

Jx  =  lUo.  g^j^  q£  ^Yie  two  numbers  is  equal  to  9aj. 

X  =    12,  less  ;  But  by  the  conditions  of  the  example. 

Sx  =    9C,  greater.         the  sum  of  the  two  numbers  is  equal  to 

108  ;  hence,  9a;  is  equal  to  108  ;  and  x, 

the  less   number,  must  be  J  of  108,  or  12;   and   8a;,  the  greater 

number,  must  be  8  times  12,  or  96. 

4.  The  greater  of  two  numbers  is  6  times  the  less,  and  tbeir 
sum  is  147  ;  what  are  the  numbers  ? 

Ans,  Less  number,  21 ;  greater  number,  126. 

5.  A  and  B  together  had  $100,  and  B  had  3  times  as  much 
money  as  A  ;  how  many  dollars  had  each  ? 

Ans,  A,  $25 ;  B,  $75. 

6.  A  man  divided  90  cents  between  two  beggars,  giving  the 
second  four  times  as  much  as  the  first ;  what  did  he  give  to 
each  ?  Ans.  First,  18  cents  ;  second,  72  cents. 

7.  Two  men,  A  and  B,  trade  in  company  with  a  joint  capital 
of  $900,  of  which  B  put  in  five  times  as  much  as  A  ;  how  much 
did  each  man  put  in  ?  Aiis.  A's  $150  ;  B's  $750. 

8.  In  a  mixture  of  720  bushels  of  grain,  there  is  three  times 
as  much  corn  as  wheat ;  how  many  bushels  of  each  ? 

^^^    (Wheat,  180  bu. 
(Corn,     540  bu. 

9.  I  expended  $12570  in  the  purchase  of  a  house  and  lot, 
and  the  house  cost  twice  as  much  as  the  lot ;  what  did  I  give 
forelch?  .       |Lot,      $1190. 

'  (House,  $8380. 

10.  A  and  B  engage  in  trade  with  joint  capital,  of  which  B 
owns  four  times  as  much  as  A.  They  gain  $7500  ;  what  is 
each  man's  share?  .    .    (A's  share,  $1500. 

^^*  IB's  share,  $6000 

SO.  1.  A  man  paid  24  dollars  for  a  hat,  vest,  and  coat. 
The  vest  cost  twice  as  much  as  the  hat,  and  the  coat  cost 
three  times  as  much  as  the  hat ;  what  was  the  pric-e  of  each  ? 


DEFINITIONS  AND  NOTATION.  Jg 

OPERATION.  Analysis.     We  represent  the 

Let    X  =  price  of  the  hat ;  price  of  the  hat  by  aj ;  as  the  vest 

2.x  =  price  of  the  vest;  ^^^^  ^^^^^  ^^  ^"^^  ^^  *^^^  ^'^^'  -^ 

Sx  =  price  of  the  coat.  '''^^   represent   the  price  of  the 

r; f- vest;  and  as  the  coat  cost  three 

6x  =  24  dollars.  ^-jj^^g  ^^  ^^^^^  ^^g  ^^^  ^^^^  3^  ^• 

X  =    4,  nat ;  represent  the  price  of  the  coat, 

2x  =     8,  vest ;  Since  x  and  2x  and  3a:  are  equa' 

Sx  =  12,  coat,  to   6x,  the  -whole   suit    cost    Gac, 

But  by  the  conditions  of  the  ex- 
ample, the  whole  suit  cost  24  dollars  ;  hence,  6a:  are  equal  to  24  dol- 
lars ;  and  a;,  the  price  of  the  hat,  must  be  |  of  24  dollars,  or  4  dol- 
lars ;  and  2x,  the  price  of  the  vest,  must  be  2  times  4  dollars,  or  8 
dollars ;  and  3x,  the  price  of  the  coat,  must  be  3  times  4  dollars,  or 
12  dollars. 

2.  A  purse  of  108  dollars  is  divided  among  three  men ;  A 
takes  a  certain  sum,  B  takes  three  times  as  much  as  A,  and  C 
takes  five  times  as  much  as  A ;  what  is  each  man's  share  ? 

{A's,  12  dollars. 
B's,  36  dollars. 
C's,  60  dollars. 

6.  A  man,  dying,  bequeathed  $30,000  to  his  wife,  son,  and 
daughter.  The  will  provided  that  the  son  should  receive  twice 
as  much  as  the  daughter,  and  the  wife  three  times  as  much  as 
the  daughter ;  what  was  the  share  of  each  ? 

r  Daughter,    $5,000. 

Ans.J  Son,  $10,000. 

(wife,  $15,000. 

4.  Divile  the  number  91  into  three  such  parts  that  the 
second  shall  be  five  times  the  first,  and  the  third  seven  times 
^'^^e  first.  (First  part,        7. 

AnsJ  Second  part,  35. 
(Third  part,     49. 

5.  Divide  the  number  96  into  four  such  parts  that  the  second 

2 


Ans. 


14  ALGEBRAIC  QUANTITIES. 

shall  be  three  times  the  first,  the  third  five  times  the  first,  and 
the  fourth  seven  times  the  first.  /  First,        6, 

.       ]  Second,  18. 

^""'■^  Third,     30. 

Fourth,  42. 

6.  A  farmer  purchased  a  certain  number  of  oxen,  three  times 
as  many  cows,  and  ten  times  as  many  sheep.  There  were  112 
in  all;  required  the  number  of  each  kind.  TOxen,     8. 

•      Ans.)  Cows,    24. 
[  Sheep,  80. 

7.  A  tax  of  $936  is  assessed  upon  four  persons,  according 
to  the  relative  values  of  their  property.  B  is  worth  three  times 
as  much  as  A,  C  four  times  as  much  as  A,  and  D  five  times 
as  much  as  A  ;  what  is  each  man's  tax  ?  /  A's,    $72. 

,  B's,  $216. 
C's,  $288. 
D's,  $360. 

8.  A  man  traveled  a  certain  distance  on  Monday,  twice  as 
far  on  Tuesday,  three  times  as  far  on  Wednesday,  and  so  on 
till  Saturday.  The  whole  week's  journey  was  504  miles ;  what 
was  the  distance  traveled  on  Monday  ?  Ans.  24  miles. 

S7.  1.  Divide  the  number  72  into  three  such  parts  that  the 
second  part  shall  be  twice  the  first,  and  the  third  part  three 
times  the  second. 

OPERATION.  Analysis.     We   represent   the 

I  ot    X  —  first  part  •  ^^^^  P^^*^  ^^  ^ '  ^^  *^^®  second  part 

2x  =  second  part ;  jf  '^"««,  f'^  Srst  part,  it  will  be 

^          +1  •    -I         +  2a;;  and  as  the  third  part  is  three 

ba;  =  third  part  ^j^^^^^  ^j^^  ^^^^^^^  ^^^,^^  j^  ^^,jjj  ^^ 

9^  =  72.  three  times  2x,  or  6a;.     The  sum 

OC  =  S,  first  part ;  of  all  the  parts,  6a:  and  2a;  and  x, 

2x  =  16,  second  part;  is  9^,  which  must  be  equal  to  72, 

Qx  =  48   third  Dart  ^^^  whole  number;  hence  x,  the 

first  part,  is  I  of  72,  or  8 ;  2a;,  the 

second  part,  i?  2  times  8,  or  16 ;  and  6a;,  the  third  part,  i;?  6  times  8, 
or  48.          '       . 


DEFINITIONS  AND  NOTATION.  15 

2.  Three  gentlemen  contributed  $500  for  a  charitable  object. 
A  gave  a  certain  sum,  B  gave  three  times  as  much  as  A,  and 
C  gave  twice  as  much  as  B  ;  what  did  each  contribute  ? 

rA,    $50. 

A71S.  3  B,  $150. 

[  C,  $:J00. 

3.  An  orchard  contains  four  times  as  many  cherry-trees  as 
pear-trees,  and  twice  as  many  peach-trees  as  cherry-trees  ;  the 
whole  number  of  trees  in  the  orchard  is  156 ;  what  is  the  num- 
ber of  each  kind  ?  C  Pear,      12. 

Ans.  <  Cherry,  48. 
^t  [Peacli,   96. 

4.  Pivide  the  number  147  into  three  such  parts  that  tlie 
second  part  shall  be  five  times  the  first,  and  the  third  part 
three  times  the  second.  (  First  part,        7. 

Ans.  <  Second  part,  35. 
I  Third  part,  105. 

5.  A  man  performed  a  journey  of  624  miles..  He  traveled 
twice  as  far  by  railroad  as  by  stage,  and  five  times  as  far  by 
steamboat  as  by  railroj^d ;  how  many  miles  did  he  travel  by 
steamboat?  '  Ans.  480  miles. 

6.  A  man  cancelled  a  debt  of  $873  by  paying  a  certain  sum 
on  Monday,  twice  that  sum  on  Tuesday,  three  times  Tuesday's 
payment  on  Wednesday,  four  times  Yi^ednesday's  payment  on 
Thursday,  and  so  on  till  Saturday ;  what  sum  was  paid  on 
Monday?  A^ns.  $1. 

7.  A  mere]] ant  gnined  three  times  as  much  by  his  business 
the  second  year  as  the  first,  as  much  the  third  year  as  tlie 
second,  and  twice  as  much  the  fourth  year  as  the  third.  Ilia 
entire  gain  was  $9750  ;  wliat  did  he  gain  the  fourth  year  ? 

Ans.  $15u0. 
@8.  1    Divide  the  number  35  into  three  such  parts  that  the 
second  shall  be  four  times  tlic  first,  and  the  third  one  lialf  the 
second. 


16 


ALGEBRAIC  QUANTITIES. 


Let 


OPERATION. 

X  =  first  part ; 
4j7  =  second  part ; 
2x  =  third  part. 
7a;  ==  85. 

X  =    b,  first  part ; 
4:x  =  20,  second  part ; 
2x  =  10,  third  part. 


Analysis.  We  represent  the  first 
part  hy  a; ;  as  the  second  part  is  4 
times  the  first,  it  will  be  4a; ;  and  as 
the  third  part  is  one  half  the  second, 
it  will  be  one-half  of  Ax,  or  2x,  The 
sum  of  all  the  parts,  2x  and  Ax  and  x, 
is  Ix,  which  must  be  equal  to  35,  the 
whole  number ;  hence  x,  the  first 
part,  is  4  of  35,  or  5  ;  Ax,  the  second 
part,  is  4  times  5,  or  20;  and  2a;, 
the  third  part,  is  2  times  5,  or  10. 

2.  A,  B,  and  C,  together  have  104  dollars,  B  has  nine  times 
as  much  as  A,  and  C  has  one  third  as  much  as  B ;  what  num- 
ber of  dollars  has  each  I  TA,  $8. 

Ans.  \  B,  72. 
[C,  24. 
8.  There  are  three  trees  which  together  bear  32  bushels  of 
apples.     The  second  bears  twelve  times  as  many  as  the  first, 
and  the  third  one  fourth  as  many  as  the  second ;  ho\Y  many 
bushels  does  the  first  bear  ?  Ans.  2  bushels. 

4.  The  sum  of  four  numbers  is  510.  The  second  is  six  times 
the  first,  the  third  is  three  times  the  second,  and  the  fourth  is 
one  half  the  third  ;  what  is  the  fourth  number  ?    Ans.  135. 

5.  Four  men  together  are  taxed  480  dollars.  B's  tax  is 
four  times  A's,  C's  tax  is  six  times  B's,  and  D's  tax  is  one 
eighth  of  C's  ;  what  is  C's  tax  ?  Ans,  360  dollars. 

SS,  1.  John  had  a  certain  number  of  marbles,  James  had 
three  times  as  many  as  John,  and  William  had  as  many  as  both 
John  and  James ;  they  all  had  64  ;  how  many  had  each  ? 

Analysis.     We  let  x  repre- 
sent the    number    of    marble 
John  had ;  as  Janies  had  three 
times  as  many,  3a;  will  represent 
his   number;   and   as   William 


Let 


OPERATION. 

57  =  J  ohn's  number ; 
ox  =  James's  number ; 
4:x  =  William's  number. 


Sx  =  64. 
X  =    S,  John's  ; 
i^x  =  24,  James's  ; 
Aujc  «=  32,  William's 


had  as  many  as  both  John  and 
James,  x  added  to  3a;,  or  4a;, 
will  represent  William's  num 
ber.    Axy  and  3x,  and  a;,  are  8a;, 
which   represents   the   numboi 


DEFINITIONS  AND  NOTATION.  17 

they  all  had  but  by  the  conditions  of  the  problem,  they  all  had  64 ; 
hence,  8a;  are  equal  to  64 ;  and  x  is  J  of  64,  or  8  ;  3a;  is  three  times  8, 
or  24 ;  and  4a;  is  four  times  8,  or  32. 

2.  Divide  the  number  100  into  three  such  parts,  that  the 
second  part  shall  be  four  times  the  first,  and  the  third  part  as 
much  as  the  sum  of  the  first  and  second.  f  First,      10. 

AnsJ  Second,  40. 
(Third,     50. 

3.  Divide  the  number  105  into  three  such  parts,  that  the 
second  part  shall  be  four  times  the  first,  and  the  third  part 
twice  the  sum  of  the  first  and  second.  ("First,        7. 

Ans.J  Second,  28. 
(Third,     70. 

4.  Four  men  together  contribute  $5250  to  build  a  parson- 
age. A  gives  a  certain  sum,  B  gives  three  times  as  much  as 
A,  C  gives  three  times  as  much  as  A  and  B  together,  and  D 
gives  one  third  as  much  as  B  and  C  ;  how  much  does  A  con- 
tribute ?  Ans.  $250. 

5.  A  man  gave  $324  for  a  horse,  carriage,  and  harness. 
The  horse  cost  five  times  as  much  as  the  harness,  and  the  car- 
riage cost  one  half  as  much  as  both  horse  and  harness  ;  what 
was  the  price  of  each  ?  C  Harness,     $36. 

Ans.  }  Horse,       $180. 
[  Carriage,  $108. 

6.  Divide  the  number  1008  into  three  such  parts,  that  the 
second  part  shall  be  nine  times  the  first,  and  the  third  part 
one  fifth  of  the  sum  of  the  other  two.  C  First,        84. 

Ans.  }  Second,  756. 
[  Third,     168. 

7.  What  number  is  that,  which  being  multiplied  by  7,  and 
the  product  added  to  the  number,  the  sum,  product,  and  given 
number  will  together  be  equal  to  80  ?  Ans.  5. 

8.  A  grocer  has  four  kinds  of  weights.  One  of  the  second 
kind  will  balance  two  of  the  first,  one  of  the  third  kind  will 
balance  two  of  the  second,  one  of  the  fourth  kird  will  ba- 

2*  "  B 


18  ALGEBRAIC  QUANTITIES. 

lance  two  of  the  third,  and  one  of  each  kind,  placed  together 
in  the  scales,  will  balance  15  pounds ;  what  are  the  denomina- 
tions of  these  weights  ? 

Ans.  1  pound,  2  pounds,  4  pounds,  and  8  pounds 

30.  1.  There  are  three  numbers,  whose  sum  is  96.  The 
second  is  four  times  the  first,  and  the  third  is  equal  to  the  first 
subtracted  from  the  second ;  what  are  the  numbers  ? 

^^^^  .  r^^rs^T  Analysis.  We  represent  the  first  by 

OPERATION.  .        ^,  1  -    A  ,'  .^      n    : 

x;  since  the  second  is  4  times  the  first, 

XjCj     X  =^  nrst ;  it  will  be  4x  ;  and  as  the  third  is  equal 

Ax  =  second  ;  ,  to  the  first  subtracted  from  the  second, 

8^  =  third.  it  will  be  4x  minus  x,  or  3a;.     The  sum 

g^  __  9Q  .  of  all  the  numbers,  3a;,  4a;,  ?nd  x,  is  8a;, 

a;  =  12    first  •  which  is  equal  to  96  ;  hence  x,  the  first 

ix  -  48,'  second ;  °""''^f '. ''  f.  '^^  ^^'  °'  }^ '  ^'  *« 

f,^    ,,  .   ,  second,  is  4  times  12,  or  48;  and  3a;, 

3.r  =  ob,  third.  xi    ..i  •  ^  •   o  i.-        to       o^ 

^  the  third,  is  3  times  12,  or  36. 

2.  The  sum  of  three  numbers  is  108.  The  first  multiplied 
by  7  will  give  the  second,  and  three  times  the  first  taken  ^rom 
the  second  wdll  leave  the  third  ;  what  are  the  numbers  ? 

Ans.  First,  9  ;  second,  63 ;  third,  86. 

3.  John  has  a  certain  number  of  marbles,  James  has  five 
times  as  many  as  John,  Henry  has  as  many  as  twice  John's 
subtracted  from  James's,  and  Henry's  added  to  John's  are 
equal  to  20  ;  how  many  has  each  ? 

Ans.  John,  5  ;  James,  25 ;  Henry,  15. 

4.  Divide  the  number  8488  into  three  such  parts,  that  the 
second  part  shall  be  five  times  the  first,  and  the  third  part  one 
half  the  difference  of  the  first  and  second. 

.4ns.  First,  436  ;  second,  2180  ;  third,  872 

5.  The  difference  of  two  numbers  is  12  times  the  less  num 
ber,  and  5  times  the  less  number  subtracted  from  the  greatei 
is  equal  tc  120  ;  what  are  the  numbers  ? 

Ans  {^^^^'     ^^' 
I  Greater,  195. 


DEFINITIONS   AND   NOTATION.  19 

31,  1.  A  bookseller  being  asked  tlie  value  of  two  pencils, 

answered  that  the  second  cost  twice  as  much  as  the  first,  and 

that  his  price  mark,  a,  would  represent  the  cost  of  the  two ; 

what  was  the  cost  of  each  ? 

OPERATION.  Analysis.     We  let  x  repre- 

Let    w  -  cost  of  the  first ;  ««'^'  *''«  ''"^'  "*  '^'^/''^^  ?«"«'•' 
o                J.    X?  ±\.               1  and,  as  the  second  cost  twice 
zx  -  cost  of  the  second.  1,0       -n  v    ^i 
as  much,  Zx  ■will  be  the  cost 

da;  =.  a  of  the   second.     The  cost  of 

a     r.  ^i  ,  both  is  2x  added  to  ar,  which 

3'       ^   '  is  3x;  but  by  the  conditions  of 

2^  the  example  the  cost  of  both 

2x  =  -^i  second  -was  a ;  hence  3a;  are  equal  to 

a ;  x,  the  cost  of  the  first,  is  J 

of  a,  or  --  ;  and  2x  is  two  times  ^-,  or  ^-. 

o  '  00 

Note. — The  expression  — -  indicates  the  division  of  the  number  a  by 

o 

3,  and  is  read,  a  divided  by  8. 

2.  If  in  the  example  above,  the  letter  a  represent  9  cents, 
how  many  cents  is  each  pencil  worth  ? 

Ans,  First,  3  cents ;  second,  6  cents. 

8.  A  certain  number,  represented  by  the  letter  c,  is  divided 
into  two  such  parts  that  the  greater  is  four  times  tlie  less ; 

what  are  the  parts  1  I  t^  c 

*  Less, 


Ans. 


5* 
4c 


Greater,   - 
5 


4    Divide  the  number  m  into  three   such  parts,  that  the 

second  part  shall  be  twice  the  first,  and  the  third  part  three 

times  the  first,  /  -r^.    .        '^^^ 

First,      -— 

D 

2m 
Ans,  {  Second,  -77-. 
'  6 

^  Third,     -^. 

5.  Divide  the  number  n  into  four  such  pan»,  that  the  second 
part  shall  be  twice  the  first,  the  third  part  as  mucl\  as  the  first 


Ans.  . 

I  Brandy, 


20  ALGEBRAIC  QUANTITIES. 

and  second;  and  the  fourth  part  as  much  as  the  first,  second^ 
and  third. 

*"      Ans.  First,  ^ ;  second,  ^^  ;  third,  ^-^  ;  fourth,  :r^. 

6.  B  is  three  times  as  old  as  A,  and  C  is  four  times  as  old 
as  A  ;  the  sum  of  their  ages  is  d ;  what  is  the  age  of  C  ? 

Ans.   -^-, 

7.  A  cask  which  held  b  gallons  was  filled  with  a  mixture  of 
brandy  and  water,  and  there  was  ten  times  as  much  brandy  as 
water;  how  much  was  there  of  each  ? 

j  Water,     ^ ; 

106 
11' 

S2,  The  following  examples  do  not  require  for  solution  the 
use  of  equations ;  they  are  given  to  exercise  the  pupil  in  al- 
gebraic notation,  and  embrace  only  given  relations  of  known 
quantities. 

1.  If  the  number  of  weeks  in  a  year  be  represented  by  m, 
what  will  express  the  number  of  days  ? 

Analysis.  There  are  7  days  in  one  week,  and  in  m  weeks  there 
must  be  m  times  7  days,  or  7m  days,  the  answer. 

2.  A  man  labored  b  days  at  c  dollars  a  day ;  how  much 
wages  did  he  earn  ?  Ans.  be  dollars. 

3.  John  had  m  m.arbles,  which  he  sold  for  m  cents  apiece ; 
now  many  cents  did  he  receive  for  them  ?        Ans.  vf  cents. 

4.  An  orchard  contains  b  rows,  in  each  row  are  c  trees, 
and  each  tree  bears  d  bushels  of  apples ;  how  many  bushels 
of  apples  does  the  orchard  produce  ?  Ans.  bed. 

5  The  height  of  a  rectangular  box  is  h,  the  breadth  is  b, 
and  the  length  is  I ;  what  are  the  solid  contents  ? 

A71S.  hbl 

6.  John  has  m  marbles,  James  has  n  times  as  many  as  John, 
and  William  has  n  times  as  many  as  James  ;  how  many  marbles 
has  William  ?  Ans,  mn^. 


DEFINITIONS  AND  NOTATION.  21 

7.  A  prize  v^^as  divided  among  three  men.  The  first  mao 
received  a  dollars,  the  second  man  h  dollars,  and  the  third  man 
c  times  as  many  dollars  as  the  other  two  ;  how  many  dollars 
did  the  third  man  receive  ?  An^.  (a  -J-  ^)^- 

8.  A  man  purchased  three  books.  For  the  first  he  gave  a 
dollars,  for  the  second  h  times  as  many  dollars  as  for  the  first, 
and  for  the  third  c  times  as  many  dollars  as  for  the  other  two  j 
vfhat  did  he  give  for  the  third  ?  An^,   {a  +  ab)G. 

9.  A  man  started  in  business  with  a  capital  of  c  dollars.  The 
first  year  he  doubled  his  money,  the  second  year  he  gained  b 
times  his  first  capital,  the  third  year  he  lost  d  dollars,  and 
dying,  left  his  property  to  n  children;  how  much  did  each 

receive  ?  .        2g  -{•  he  —  d 

Ans, . 

n 

10.  Four  men  having  joint  partnership  in  a  nursery,  sold  m 

rows  with  m  trees  in  a  row,  at  m  cents  apiece  ;  the  trees  were 

packed  in  h  bunches,  and  delivered  at  an  expense  of  h  cents  a 

bunch  ;  what  did  each  man  realize  by  the  sale  ? 

m^ — li^ 
Ans.  — J — . 

33.  To  find  the  numerical  value  of  an  algebraic  quantity, 
-vhen  the  literal  factors  represent  known  numbers,  we  must 
substitute  the  given  numbers  for  the  letters,  and  perform  upon 
them  the  operations  indicated. 

Til^  — —  b^ 

1.  What  is  the  numerical  value  of  — j — ,  when  m  =  40, 
and  6  =  8? 

OPERATION.  Analysis.     By 

m»--  6'  _  40  X  40  --  8  X  8  __  the  conditions  of 

~J        —  J  —  ^'^'*  the  example,  7h^  is 

equal  to  40  X  40, 
or  1600 ;  S^  is  equal  to  8  X  8,  or  64 ;  and  1600,  less  64,  is  1536,  which, 
divided  by  4,  gives  384,  the  answer. 

Find  the  numerical  values  of  the  following  algebraic  expres- 
sions, in  each  of  which 

a  =  12  5C=«10;m  =  8;n  =  5 


22  ALGEBRAIC  QUANTITIES. 


2 

ao  —  m\ 

3. 

(a  +  c)m. 

4. 

{a?—-am)n. 

5. 

(a  +  G  -{■  m  +  n)am. 

6 

a^  —  mn 
4       • 

7. 

am  +  'inhi  —  2m 

mn 
(a^  —  cmn)m 


a 


Ans.  56. 

Ans.  176. 

Ans.  240. 

.4ns.  3360. 

Ans.  26. 

Jns.  10. 

Ans.  885^. 


9. ; ^-.  Ans.  32. 

(a  +  c  +  m)7i 

10.  7^^ — ^ — , — -^-— .  ^ws.  6. 
(a  +  c)?i  4-  en 

11.  ('^-^  +  ^-"^)«\  ^„s.  27. 

7ac~-m'^ 

12. -.  Ans.  di. 

ba  —  nv 

13.  4 (3a  -f  2m).  Ans.  208., 

^ ,     5m(a^  4-  'rnc  —  14m)  .        ^ 

14.  — ^^ — :; ^.  Ans.  7. 


15.  — r, -, — 5  ^^^s.  1775. 

m^  —  2mn  +  n  ^ 

16.  ac ^^ -^  Ans.  113. 

a 

^^       am          3a  >i        10 

17. 1 .  Ans.  12 

a  +  m        71 

18.  ?£(^J!^.  Jns.  456. 

Ibn 

19.  /l!!^_^)^i.  ^ns.  16. 
\  n         0/ 

20.  (a  +  c)  (m  +  n).  ^Itis.  286. 

21.  -^^- n.  ^ns.  15. 

m  4-  2n 

22.  fl2  _|.  ^2  —  c'  —  n».  ^n8..  83. 


TERMS.  23 

TERMS. 

34«  The  Terms  of  an  algebraic  quantity  are  the  divisions 
made  by  the  signs,  +  and  — ;  thus,  in  the  quantity  oa+¥ —  mx^ 
there  are  three  terms,  of  which  3a  is  the  first,  +  ¥  is  the  second^ 
and  —  7nx  is  the  third. 

3^.  Positive  Terms  are  those  which  have  the  plus  sign  ; 
as,  +  a  or  +  b'^d.  The  first  term  of  an  algebraic  quantity,  if 
written  without  any  sign,  is  positive,  the  sign  +  being  under* 
stood. 

36.  Negative  Terms  are  those  which  have  the  minus  sign  ; 
as,  —  2a,  or  —  3c^d  The  sign  of  a  negative  quantity  is  never 
omitted. 

37.  Similar  Terms  are  terms  containing  the  same  letters, 
affected  with  the  same  exponents ;  the  signs  and  coefficients 
may  differ,  and  the  terms  still  be  similar.  Thus,  Ca^  and' 5a^ 
are  similar  terms ;  Ih'd  and  —  bhi^d  are  similar  terms. 

38.  Dissimilar  Terms  are  those  which  have  different  letters 
or  exponents ;  thus,  ahc  and  acd  are  dissimilar  terms ;  ax^y^ 
and  d^xy  are  dissimilar  terms. 

39.  A  Monomial  is  an  algebraic  quantity  consisting  of  only 
one  term  ;  as,  4a,  ocd,  or  *lh^x. 

4©.  A  Polynomial  is  an  algebraic  quantity  consisting  of 
more  than  one  term  ;  a  +  6  ;  or  odb  —  2x  -\-  c. 

41.  A  Binomial  is  a  polynomial  of  two  terms ;  as  a  +  c, 
or  2.x  —  y. 

4S.  A  Eesidual  is  a  binomial,  the  two  terms  of  which  are 
connected  by  the  minus  sign  ;  as,  a  —  b,  or  ilx  —  2y. 

43.  A  Trinomial  is  a  polynomial  of  three  terms  ;  as  x  -{- 
y  -{-  z,  or  2a  —  26  -f  c\ 

44.  The  Degree  of  a  terra  is  the  number  of  literal  factors 
it  contains,  and  is  found  by  adding  the  exponents  of  the  several 
letters;  thus,  a  and  S6  are  terms  of  the  first  degree;  a^  and 

Define  the  terms  of  an  algebraic  quantity.  Positive  terms.  Negative 
terms.  Similar  terms.  Dissimilar  terms.  A  Monomial.  A  Polynomial. 
A  Binomial.     A  Residual.     The  Degree  of  a  term. 


24  ALGEBUAIC  QUANTITIES. 

2ab  are  terms  of  the  second  degree ;  a^  Sa-b,  and  bahc  ard 
terms  of  the  third  degree. 

45.  A  Homogeneous  quantity  is  one  whose  terms  are  all 
of  the  same  degree ;  as,  oc^  —  8a;^2/  +  xyz, 

AXIOMS. 

46.  An  Axiom  is  a  self-evident  truth. 

The  principles  of  all  algebraic  operations  are  based  upon 
the  following  axioms : 

1.  If  the  same  quantity  or  equal  quantities  be  added  t:> 
equal  quantities,  their  sums  will  be  equal. 

2.  If  the  same  quantity  or  equal  quantities  be  subtracted 
from  equal  quantities,  the  remainders  will  be  equal. 

3.  If  equal  quantities  be  multiplied  by  the  same,  or  equal 
quantities,  the  products  will  be  equal. 

4.  If  equal  quantities  be  divided  by  the  same,  or  by  equal 
quantities,  the  quotients  will  be  equal. 

5.  If  the  same  quantity  be  both  added  to  and  subtracted- 
from  another,  the  value  of  the  latter  will  not  be  altered. 

6.  If  a  quantity  be  both  multiplied  and  divided  by  an- 
other, the  value  of  the  former  will  not  be  altered. 

7.  Quantities  which  are  respectively  equal  to  any  other 
quantity  are  equal  to  each  other. 

8.  Like  powers  of  equal  quantities  are  equal. 

9.  Like  roots  of  equal  quantities  are  equal. 

10.  The  whole  of  any  quantity  is  greater  than  any  of  ite 
parts. 

11.  The  whole  of  any  quantity  is  equal  to  the  sum  of  all  its 
parts. 

Define  a  Homogeneous  Quantity.    An  Axiom.     Repeat  the  Axioms 
giveiL 


ADDITION.  25 


ADDITION. 

47.  Addition,  in  Algebra,  is  the  process  of  uniting  iwo  or 
more  quantities  into  one  equivalent  expression,  called  their 

sum. 

Since,  in  algebra,  the  quantities  to  be  added  may  be  either 
positive  or  negative,  it  is  necessary  to  consider  here  more  fully 
the  nature  of  the  signs  +  and  — .  Thus  far  they  have  been 
employed  to  indicate  simply  the  opposite  processes  of  addition 
and  subtraction.  They  have,  however,  a  wider  significance, 
and  indicate  not  only  operations  to  be  performed^  but  the 
quality,  or  relative  character  of  the  quantities  to  which  they 
are  applied.  They  may  denote  opposite  directions  in  space, 
opposite  effects  in  nature,  or  opposite  results  in  business. 
Thus,  if  plus  indicate  direction  north,  minus  will  indicate 
direction  south ;  if  plus  indicate  heat,  minus  will  indicate 
cold  ;  and  if  plus  indicate  gain,  minus  will  indicate  loss. 

CASE   I. 

48.  To  add  similar  terms. 

1.  A  cooper  made  7  barrels  on  Monday,  9  barrels  on  Tues- 
day,  and  6  barrels  on  Wednesday ;  how  many  barrels  did  hi  j 
make  in  the  three  days  ? 

AKITHMETICALLY. 

7  barrels.  . 

9  barrels. 
6  barrels. 
Ans,  22  barrels. 

Define  Addition.  Explain  the  nature  of  the  signs  -f  and  — .  What 
is  Case  I  ?    Give  Analysis. 

3 


26  ENTIRE  QUANTITIES. 


ALGEBRAICALLY.  Analysis.    We  represent  1  barrel  b;^  fche 

fjy  letter  b ;  then  76  will  represent  7  barrels,  96 

will  represent  9  barrels,  and  66  will  represent 

^  6  barrels ;  and  since  7  barrels,  9  barrels,  and 

^  6  barrels  are  22  barrels,  76,  96,  and  66  are 

22b  226. 


2.  A  mass  of  iron  and  wood  is  submerged  in  water  by  its 
own  gravity.  The  iron  tends  to  sink  with  a  force  of  20 Z,  and 
the  wood  buoys  upward  with  a  force  of  16 Z ;  what  will  the 
whole  mass  weigh  while  under  water  ? 

^■.^^,>  *  m-r^T^r  Analysis.    Wb   Indlcate  actual  weight  by  the 

OPERATION.         ,        .  ,     ,  .       /»  ,  , 

plus  sign,  and  the  opposite  force,  or  buoyancy,  by 

•  the  minus  sign.     Since  the   tendency  to   sink  is 

'ZL—L  greater  by  4Z  than  the  tendency  to  rise,  the  mass 

4-    4Z  has  a  weight  of  4Z ;  hence,  +  20Z  and  —  16Z  united 

are  +  4Z. 

Note.  —  The  answer,  -}-  4Z,  in  the  above  example,  is  called  the  alge- 
braic sum  of  the  two  forces,  +  20Z  and  —  16Z,  because  it  shows  their 
united  effect, 

3.  A  ship  started  at  the  equator  and  sailed  the  first  day  16 
miles  north,  the  second  day  20  miles  south,  the  third  day  8 
miles  north,  and  the  fourth  day  7  miles  south ;  how  far  from 
the  equator,  and  in  what  latitude,  was  the  ship  at  the  end  of 
the  four  days  ? 

FIRST  OPERATION  ANALYSIS.    Weletwrepre- 

j^  l^T^  —  20m  sent  1  mile ;  and  to  distinguish 

I      3^ 7^  the  directions,  we  indicate  dis- 

-— ^-j jj=—  tance  north  by  the  plus  sign, 

+  24m-27m  =  -3m  and   distance    south  by  the 

minus  sign,  writing  the  positive  terms  in  one  column  and  the  nega* 
tive  terms  in  another  column.  The  whole  distance  sailed  north,  is 
+  ICm  and  +  8m,  which  is  +  24m ;  and  the  whole  distance  sailed 
south,  is  —  20m  and  —  7m,  which  is  —  27m ;  and,  as  27  is  3  more 
than  24,  the  ship  must  be  three  miles  south  of  the  equator,  ex 
pressed  algebraically  thus,  —  3m. 

Explain  the  diflference  between  Arithmetical  and  Algebraic  Addition, 
S8  shown  in  example  3. 


ADDITION.  27 

SECOND  OPERATION.  Analysis.  In  the  second  operation  we 
+  16m  write  all  the  terms  in  one  column,  since 
—  20m  they  are  similar.  +  16m  and  -f  8m  are 
4-  8m  +  24m ;  and  —  20m  and  —  7m  are  —  27m ; 
__  7^  and  +  24m  —  27m  are  —  3m,  the  algebraic 
o —  sum  of  the  quantities  in  the  column. 

Note. — There  is  a  dis%ction  between  arithmetical  and  algebraic  addi- 
tion. In  the  above  exanl^^le  had  the  question  been,  how  far  the  ship 
sailed,  the  answer  would  be  16  -|-  20  +  8  -|-  7  =  51  miles,  or  51  w,  which 
is  the  arithmetical  sum  of  the  distances  sailed.  But  the  real  question  is, 
what  distance^  north  or  south,  did  tJie  ship  make  from  the  point  of  starting  ; 
and  the  answer  is  3  miles  south,  or  —  Sw,  which  is  the  algebraic  sum  of 
the  distances  sailed.  Hence,  adding,  in  algebra,  does  not  always  aug- 
ment. Positive  and  negative  quantities  represent  things  opposite  in 
kind  or  quality,  and,  if  similar  in  denomination^  are  added  or  united,  by 
apparent  subtraction. 

From  these  examples  and  illustrations  we  derive  the  fol- 
lowing 

Rule.  I.  Wlien  the  signs  are  alike,  add  the  coefficients, 
and  prefix  the  sum  with  its  proper  sign  to  the  common  literal 
part, 

II.  When  the  signs  are  unlike,  find  the  sum  of  the  posi- 
tive and  of  the  negative  coefficients  separately,  and  prefix  the 
difference  of  the  two  sums  with  the  sign  of  the  greater,  to  the 
common  literal  part, 

EXAMPLES  FOR  PRACTICE. 


(4.) 

(5.) 

(6.) 

(7.) 

(8.) 

3a 

2m' 

—  Sbx 

—  4a'6o 

-h  7cd^ 

9a 

6m' 

—  56a; 

—  5a'6c 

+  Scd:" 

5a 

5m' 

—  46a; 

—  12a'6o 

+  2cd^ 

12a 

10m' 

—  26a; 

—  a^6c 

+     cd^ 

a 

5m' 

—  76a; 

—  14a'6c 

+    Q^cd:" 

2a 

7m' 

—  bx 

—  2a'6c 

+    4c(? 

32a 

t^^ 

—  226a; 

—  -  ■  "  ^  ■  ■ 

+  23ceP 

Give  the  rule  for  the  algebraic  addition  of  similar  terms. 


28  ENTIRE  QUANTITIES. 


(9.) 

(10.) 

(11.) 

(12.) 

(13.) 

—  5a 

+  3aa?^ 

+ 

Sar" 

—   5a' 

+  36y 

+  4a 

+  4aa7^ 

— 

6ar' 

—  Ida? 

+  96y 

+  6a 

—  '^ax' 

— 

IQaf 

4-lOa^ 

—  i06y 

—  3a 

—  ^ax" 

+ 

3x' 

+  Ua? 

_  i9&y 

+    a 

4-5aa72 

+ 

2a? 

+    Qa' 

_  26y 

+  oa 

—  2ax^ 

— 

^a? 

^15.' 

—  i96y 

(14.) 

(15.) 

(16.) 

(17.) 

bh'd 

2mn 

—  25a'6o 

147z» 

•^Wd 

4:mn 

ZWbo 

—    25z» 

Wd 

— . ' 

I2mn 

—  na^ho 

123» 

—  Wd 

16mw 

48a'6o 

—     143» 

18.  What  is  the  sum  of  57?i,  7w,  11m,  m,  and  12m?  "t 

-4ns.  36m. 

19.  What  is  the  sum  of  4a,  7a,  6a,  and  10a  ? 

^ns.  27a. 

20.  What  is  the  sum  of  —  5c^  l^c\  —  28c^  and  —  c"  ? 

Ans.  —  24c\ 

21.  What  is  the  sum  of  —  1256e,  —  1686c,  and  26c? 

Ans.  —2916c. 

22.  What  is  the  sum  of  Sxy,  —  4^i/,  lOxy^  and  —  7xy  ? 

Alls,  2xy. 

23.  What  is  the  sum  of  9a'5c,  3a'6c,  —  8a'6c,  —  2a-6c, 
5a*^6c,  and  —  15a^6c  ?  Ans.  —  8a^6c. 

24.  Add  dni^Xf  —  2m^Xf  —  7m^a7,  and  28m^5?. 

Ans.  24m^x, 

25.  Add  7^y,  —  15:ry,  +  12a;^?/^  —  6xy,  and  xy, 

Ans,  —  xy. 

In  the  same  manner  similar  quantities,  of  whatever  kind, 
may  be  added  by  taking  the  algebraic  sum  of  their  coeffi- 
cients.    The  pupil  will  observe  that,  since  2  times  any  num- 


ADDITION.  29 

ber  whatever,  added  to  three  times  the  same  number,  are  5 
times  that  number,  so  2(a  +  ^),  added  to  3(a  -f  6),  are  5(a  +  &). 


(26.) 

(27.)                 (28.) 

(29.) 

^c-x) 

(x+y)     _4(2a-6) 

4(ar  — 2/  +  3) 

7(c  —  x-) 

—  3(a;  +  2/)     _7(2a  — 6) 

7(a;— y+3) 

lOic  —  x) 

20(x4-y)          8(2a_6) 
18(a;#>i/)      _3(2a  — 6) 

—  12Cx—y  +  3) 

n(c  —  x) 

—  ix-y  +  B) 

(30.)                           (31.) 

(32.) 

8a(a 

+  6)           7(Qx  +  y  —  zy 

4(62/  +  &) 

7a(o 

+  b)        -SiQx  +  y  —  zy 

—  3(6y  +  6) 

—  5a(a 

+  &)       _  2(6x4- 3/  — 2;)' 

7(62/  +  6) 

Sa(a 

+  6)            3(6a;  +  2/  — z)'' 

-2(62/ +  6) 

83.  What  is  the  sum  of  S(z  —  m),  b(^z  —  m),  —  12(z  —  m), 
_-  14(z  —  m),  and  10(2  —  m)  ?  ^ns.  —  8(2  —  m). 

84.  What  is  the  sum  of  —  4(a  -r  2&)^  5(a  4-  26)^ 
^  12(a  +  2by,  and  20(a  +  2by  ?  Ans.  9(a  +  26)1 

85.  What  is  the  sum  of  {x  +  1),  5(x  +  1),  8(0?  +  1),  and 
—  8(^  +  1)?  Jws.  (x  +  1). 

CASE  n. 

49.  To  add  polynomials. 

It  is  evident  that  dissimilar  terms  may  be  added  by  writing 
them  one  after  another,  connected  by  their  proper  signs ;  but, 
if  in  the  same  polynomial,  or  in  the  different  polynomials  to  be 
added,  there  be  similar  terms,  these  may  be  united  by  Case  I, 
and  a  reduced  expression  obtained ;  and  it  is  immaterial  in 
wliat  order  the  terms  are  written  in  the  aggregate  sum,  since 
the  whole  is  equal  to  the  sum  of  all  its  parts,  in  whatever  order 
the  parts  are  taken,  (Ax.  11). 

1.  Add  Ba  +  2hc,  4a  —  7bc  +  m,  and  x^  —  2a  +  36c. 

What  is  Case  II? 
8* 


90  ENTIRE  QUANTITIES. 

Analysis.     We  write   3a,  4a, 

OPERATION.  ^^^  —  2a,    in    one    column,   be- 

8a  +  2bc  cause  they  are  Bimilar  terms,  and 

4^ *jIq  ^  yfl  2bc,  —  7dc,   and  36c,  in  another 

« 9/7  4_  ^h/»  column,  for  the  same  reason ;  andx* 

_—  and  m  not  being  similar  to  each 

a;*  +  5a  —  26c  +  m  other,  or  to  any  of  the  other  terms, 

we  write  them  in  separate  columns. 
Commencing  at  the  left,  we  write  x^  in  the>«um ;  then  —  2a,  -}-  4a, 
and  -f  3a  are  +  5a,  which  we  write  under  the  column  added ;  and 
•+•  36c,  —  76c,  and  +  26c,  are  —  26c,  which  we  write  under  the 
column  added ;  finally  annexing  +  m,  we  obtain  the  entire  sum, 
a:*  +  5a  —  26c  +  m. 

From  this  example  we  deduce  the  following 

Rule.  I.  Write  similar  terms  in  the  same  column,  forming 
as  many  columns  as  there  are  dissimilar  terms  in  the  given 
quantities. 

11.  Add  each  column,  as  in  case  J,  and  connect  the  results 
by  their  proper  signs 

EXAMPLES  FOR  PRACTICE. 

(2.)                         (3.)  (4.) 

Sa^  —  Acd           4:x'y—     a'b'  5ab' —   7  do 

^x'  —  Scd           Sx"y+    9a'b'  —  m  7ab'  +  Udc 

~  5a?'  -f  ^cd  z  —  bx'y  —  12a'b'  —12ab^—   Qdc 

^x'  —  Scd  z  +  2x'y—  4a*'6'  — m  +      dc 

5.  What  is  the  sum  of  6a6  +  12bc—Scd,  Scd  —  7ab  —  9bc, 
and  12cd  —  2a6  —  56(?  ?  Ans,  7cd  —  Sab  —  26c. 

6.  What  is  the  sum  of  96*  — 3ac  +  d,  W  +  7d  —  Aac, 
M  —  46^  +  6ac,  56'  —  2ac  —  12d,  and  W  —  d? 

Ans.  lSb'  —  Sac  —  2d, 

7.  What  is  the  sum  of  7a6  —  m^  +  q,  —  4a6  —  5m'  —  3g, 
12a6  +  14m'  —  z,  and  —  6m'  —  2q  ? 

Ans.  15a6  4- 2m'  —  4^  —  z. 

Give  analysis.     Rule. 


ADDITION.  gj 

8.  What  is  the  sum  of  Ga;  —  5&  +  a  -f  8,  and  —  6a  -—i 
407  +  46  —  3?  Ans.  2x  —  b  —  4a-\-5. 

9.  What  is  the  sum  of  a  +  26  — Sc  —  lO,  36  — 4a  +  5(?  + 
10,  and  56  —  c  ?  Ans.  —  3a  +  106  +  c. 

10.  What  is  the  sum  of  3a -f  h — 10,  c  —  d  —  a,  and  —  4c  4- 
2a  — 36  — 7?  Ans,  4a  — 26  — 3c  — (i_17. 

11.  Add  15a'  —  86^c?  +  32aV  —  126c,  196V  —  4a^  + 
llaV  4-  26c,  a'^  — 29aV  — 126'^c  +  56c,  and  9aV  — 146c  -f- 
b\\  Ans.  12a'' +  23aV  —  196c. 

12.  Add  5a^6'  —  8a26'  +  x'y  +  xy\  4a'6«  —  7a«6'  —  Sxy'  + 
Qx'y,  Sa'b'  +  Ba'b'  —  Bx'y  +  dxy\  and  2a26«  — a«6^  — 3a;V— 
Sxy\  Ans.  a^b^  +  x^y. 

13.  Add  nax'  —  ^ay",  »_38a.77^  — 3ai/*  +  laf,  8  +  12ai/*, 

—  ^ay^  +  12,  and  —  34aa;*  +  bay^  —  9ay^. 

Ans.  —  2a2/»  +  20. 

14.  Add  7x^  —  5c^  +  14m^,  — Sx^  -f  icx  —  17m5r — pq^ 
4j?^  -f  12mg  +  ^pq  —  z,  2cx  —  7mg  —  2pq,  and  3a;''  —  2ca; 

—  mg  —  4pg  -f-  32.         Ans.  llo?''  —  ex  +  mg  —  4pq  +  2z. 

15.  Add  7m  +  3n  —  lip,  3a — 9n  —  11m,  Sn — 4m  +  bp, 
and  Qn  —  m  +  Sp.  Ans.  3a  —  9m  +  Sn  —  Zp, 

16.  Add  7a  —  36  +  c  -f  m,  and  36  —  7a  —  c  +  m. 

Ans.  2m, 

17.  Add  X  —  y  —  z,  and  y  —  x  +  z.  Ans.  0. 

18.  Add  3(a  4-  6),  4(a  +  b),  and  —  2(a  +  6). 

Ans.  5(a  +  6). 

19.  Add  6(m^  — nj  +  2c,  —  5(m''  — n)  +  7c,  S(m^  —  n) 

—  4c,  and  4(m''  —  n)  +  c.  Ans.  8(m^  —  n)  +  6c. 

20.  Add  2a(x  —  y'')  —  Zmz\  4a(j7  —  2/')  —  5m2;^  and  5a 
(;r  —  2/'-^)  4-  7m2;l  Ans.  lla(x  —  2/0  —  mz\ 

21.  Add  8  aa;  +  2(x  +  a)  +  36,  9aa;  +  6(x  +  a)  —  96,  and 
11a;  -1-  66  —  7aa;  —  8(a;  +  a).  Ans.  Wax  +  11a?. 


82  ENTIRE  QUANTITIES. 

•5©.  The  Unit  of  addition  is  the  quantity  whose  coefficients 
are  added ;  thus,  in  the  example,  3^  —  4^  +  7^  =  6^,  the 
unit  of  addition  is  x ;  for  the  result,  6^,  was  obtained  by  uniting 
the  coefficients  of  x  into  one  number.  Dissimilar  terms  may 
often  be  added,  by  making  some  common  letter  or  letters  the 
unit  of  addition. 

1.  What  is  the  sum  of  ax,  hx,  and  ex  ? 

Analysis.    We  make  the  conimon  let- 

OPERATION.  ter,  x,  the  unit  of  addition ;  thus  a  times 

^/j;         X,  b  times  a;,  and  c  times  x  must  be  equal 

T  to  X  multiplied  by  the  sum  of  a,  6,  and  c; 

and  since  a.  b,  and  c  are  dissimilar,  we 
ex  . 
indicate  their  addition,  inclose  th-o  sum 

Sum.  (a  +  6  +  c)x  in  a  parenthesis,  and  write  it  as  the  co- 
efficient of  X,  and  thus  obtain  the  sura 
of  the  given  quantities. 


EXAMPLES 

FOR  PRACTICE. 

(2-) 

(3.) 

(4.) 

ax 

by' 

lay 

2cx 

Bay' 

—  2ay 

Adx 

If 

—  cy 

(a  +  2c  +  4:d)x 

(b+Sa-h7)f 

(pa  —  c)y 

5.  What  is  the  sum  of  ex,  2cx  and  Gar,  when  x  is  the  unit 
of  addition  ?  Ans.  (So  +  6)^7. 

6.  What  is  the  sum  of  am^y  —  6m^  -{-  em^y  when  m^  is  the 
unit  of  addition  ?  Ans.  (a  —  6  +  c)m\ 

7.  What  is  the  sum  of  (a  H-  b)x  and  (a  H-  e)Xf  when  x  is  the 
unit  of  addition  ?  Ans,  (2a  +  6  +  c)j?. 

8.  What  is  the  sum  of  Sx,  bx,  and  (a  +  b)Xf  when  x  is  the 
unit  of  addition  ?  Ans.  (a  +  2b-{-  S)x. 

9.  What  is  the  sum  of  4:axy,  —  axy,  and  +  cxy,  when  ary 
is  the  unit  of  addition  ?  u4ws.  (3a  -f  c)^^/. 

Define  the  Unit  of  Addition. 


SUBTRACTION.  gg 


SUBTRACTION 

^S.  Subtraction,  in  algebra,  is  the  process  of  finding  the 
difi^3rence  between  two  quantities. 

CASE  I. 

53.  To  find  the  diiFerence  of  similar  terms. 

1.  A  and  B  travel  north  from  the  same  point ;  the  distance 
A  travels  is  7m,  and  the  distance  B  travels  is  47?i ;  how  much 
farther  north  is  A  than  B  ? 

OPERATION.  ^  ^  ,  ,    „     , 

Analysis.     A  must  be  as  much  farther 

Minuend,  im  j^q^^j^    ^^^j^^   jj^   ^^    g>g    distance,   4m,    sub- 

Subtrahend,         4m  tracted  from  A's  distance,  7m ;  and  7m  — 

Difference,         1^  ^m  =  3m,  the  answer. 

2  A  and  B  start  from  the  same  point ;  A  travels  north  a 
distance  of  7m,  and  B  travels  south  a  distance  of  4m ;  how 
much  farther  north  is  A  than  B  ? 

Analysis.   To  express  the  distances 

OPERATION.  algebraically,  we  indicate  the  different 

Minuend  +     7m  directions  by  opposite   signs ;   north 

.  by  pluSf  south   by  minus.     Since  A 

Subtrahend,  ^     4m  ^^^^^^^^   ^^  ^^^^^^  ^^.^^^   ^  tv^Yehd 

Difference,  +  Hm  4m   south,  A  must  be   11m  farther 

north   than  B,  and  to  indicate  his 
direction  from  B  wo  must  use  the  plus  sign,  thus  +   11m. 

The  expression,  +  11m,  in  the  last  example,  is  called  the 
algebraic  difference  of  +  7m  and  —  4m,  because  it  denotes 
their  distance  asunder.  To  subtract,  in  algebra,  is  not  in  all 
cases  to  diminish.     A  positive  and  a  negative  quantity  are  in 

Define  Subtraction.     What  is  Case  I  ? 
0 


84 


ENTIRE  QUANTITIES. 


opposite  circumstances,  or  counted  in  opposite  directions; 
he7ice  the  difference ,  or  space  between  them,  is  their  apparent 
sum.  If  we  demand  the  difference  of  latitude  between  7 
degrees  north  and  4  degrees  south,  the  answer  is  7  -}-  4  =  11 
degrees,  an  operation  which  appears  like  addition. 

From  these  two  examples,  we  learn  that  a  positive  term 
is  subtracted  by  changing  its  sign  to  minus,  and  a  negative 
term  is  subtracted  by  changing  its  sign  to  plus. 

This  principle  is  further  illustrated  by  the  example  below, 
in  which  it  is  plain  that  the  remainders  in  the  lower  line  must 
increase  by  2  throughout,  since  the  numbers  to  be  subtracted 
decrease  by  2  throughout ;  hence,  to  obtain  the  true  result, 
the  sign  of  —  2  and  — 4  must  be  changed  to  -J-. 


3.  From          16             16  16  16  16 

Take                 4                    2            0  —     2  _    4 

Reinainder,'l2"           ^ii  16  "~18  "^ 

4.  From  +  7a  subtract  -f  12a. 


OPERATION. 
Minuend,  +      7a 

Subtrahend,        -f-  12a 
Difference,  —     Ofl 


Analysis.  "VVe  change  the  sign  of 
the  subtrahend  as  in  the  other  examples ; 
then  —  12a  and  -f  7a,  are  —  5a,  the 
algebraic  difference. 


5,    From 
Take 


Difference,  10a  6a         0 


15a      10a       5a  0  Analysis.   Since 

5a         5a       5a  5a  *he    minuends    de- 

— — ■  crease  by  5a  toward 

—  ^^  the  right,  and  the 

subtrahends  are  all 

equal,  the  remainders  must  decrease  by  5a,  and  the  last  remainder 

43  therefore  —  5a. 

We  cannot,  numerically,  take  a  greater  quantity  from  a 
less,  nor  any  quantity  from  zero,  for  no  quantity  can  be  less 
^Jian  nothing.     Hence,  in  the  last  two  examples,  the  answer, 


How  is  a  positive  term  subtracted  ?    How  a  negative  term  t    What  is 
tnderstood  by  a  minus  quantity  ? 


SUBTRACTION. 


85 


—  5a,  is  not  5a  less  than  nothing,  but  5a  applied  in  the  oppo- 
site direction  to  +  5a.  To  subtract  a  quantity  algebraically, 
is  to  change  the  direction  in  which  it  is  reckoned  or  applied. 
Thus,  we  see  that  by  a  change  of  sign,  we  can  find  the  alge- 
braic difference  between  any  two  quantities  whatever. 

From  these  examples  and  illustrations  we  derive  the  fol- 
lowing 

Rule.  Change  the  sign  of  the  subtrahend,  or  conceive  it  to 
be  changed,  and  unite  the  terms  as  in  addition. 


EXAMPLES  FOR  PRACTICE. 


Prom 
Take 
Rem. 


(6.) 
+  4a 
•^    a 
+  3a 


(7.) 

+  6^ 

+  Sx" 


(8.) 
^lObc 

—  7bG 

—  36c 


(9.) 
+    4m^z 

^12^ 


Prom 
Take 
Rem. 


(10.) 

—  IQb'G 

—  llb'c 
+      b'o 


(11.) 
+  lSmd 
+  mnd 
—   2md 


(12.) 
-I-  27h' 
—  h' 
+  2Sh^ 


(13.) 

_+_2W 
—  28^» 


14.  From  llx^y  subtract  —  4:X^y,  Ans,  21x^y, 

15.  From  abed  subtract  —  abed,  Ans.  2abcd. 

16.  From  25^m  subtract  2Sgm.  Ans.  —  Sgm. 

17.  From  —  166V  subtract  46V.  Ans,  —  206W. 

18.  From  —  lls^^  subtract  — 12sq^,  Ans,  sq^ 

19.  From  SOxy  subtract  4:0xy.  Ans.  —  lOocy. 

20.  From  75mn^  subtract  —  25mnl  Ans,  lOOmn^, 

21.  From  —  75mn^  subtract  —  25mn^  Ans,  —  50mn*. 

22.  From  —  ISpqr  subtract  —  llpqr,  Ans,  — pqr, 

23.  From  Ubx^y  subtract  llbx^y,  Ans,  ^-  Sboc^y, 

24.  From  5(a  +  b)  subtract  2(a  +  6).  Ans,  3(a  -i-  6). 


How  may  the  algebraic  difference  of  any  two  quantities  be  found? 
Give  the  rule  for  the  subtraction  of  similar  terms 


86  ENTIRE  QUANTITIES. 

25.  From  la{c  —  m)  subtract  —  5a (c  — m). 

Ans.  12a(c  —  m). 

26.  From  — 11(^2  —  %j)  subtract  —  b{x'  —  y). 

Ans,  — Q(x^  —  2/). 

27.  From  12(m  —  n)  subtract  —  12(m  —  n). 

Ans,  24(m  —  n). 
28    Subtract  15^y2;  from  —  oxyz,        Ans.  —  ISxyz. 

29.  Subtract  —  Iblm^nq  from  —  l^m^nq.    Ans.  l^lm^nq, 

30.  Subtract  Vlld'hc  from  ISOa^Z^c.  Ans.  —  22a'6c. 

31.  Subtract  lipc"  —  y"—  z^)  from  12(a?2  —  2/'  —  ^'0. 

^ns.  5(^2  — 2/2 —  z^). 

32.  Subtract  12a6Q}  _  5)  from  Ibab^p  —  q), 

Ans.  Sab(j)< — q). 

33.  Subtract  m\c—  1)  from  —  2mXc  —  1). 

Ans,  —  ^m\c  —  1). 
CASE  n, 
53.  To  find  the  difference  of  polynomials. 

I.  From  a  subtract  h  —  c. 

Analysis.    We  first  subtract  6  from 

a,  and  obtain  for  a  result,  a  —  6 ;  but 

OPERATION.  ouj.  t^uQ  subtrahend  is  not  6.  but  6  —  c; 

Minuend,  a  and,  as  we  have  subtracted  a  quantity 

Subtrahend,  6 c         t^o  great  by  c,  our  remainder  must  be 

7  too  small  by  c ;  we  therefore  add  c  to 

Difference,        a  ^  b  ■{-  C  ^^^  g^^^.  ^^^^^^^  ^^^  ^^^^j^^  ^^^  ^^^^  ^^^ 

mainder,  a  —  6  +  c. 
By  this  example,  we  have  in  a  more  general  manner  estab- 
lished the  principle,  that  the  sign  of  a  term  to  be  subtracted 
must  be  changed.     Hence  the  following 

Rule.    I.  Write  the  subtrahend  underneath  the  minuend, 
placing  similar  terms  under  each  other, 

II.  Change  the  signs  of  the  terms  of  the  subtrahend^  09 
conceive  them  to  be  changed. 

What  is  Case  II?    Give  Analysis.    Rule. 


SUBTRACTION. 


37 


III.    Unite  similar  terms  as  in  addition^  and  bring  down 
all  the  remaining  terms  with  their  proper  signs. 


From 
Take 
Hem 


EXAMPLES   FOR  PRACTICE. 

(2.)  (3.) 

4a  +  2^  — .  3c  Sax  +  2y 

'^a'\-4:X — 6c  xy  —  2y 


3a  —  2j?  -4-  3c 


oax  —  xy  +  4:y 


(4.) 
b 


a 
a- 


2h 


\  From 
iTake 
•Rem. 


From 
Take 
Rem. 


From 
Take 
Rem. 


(5.) 
2x^  —  Sx  +  y^ 
—  x^  —  4:x  ■+  a 

dx^  +    X  +  y^  —  a 


(6.) 

7a  +  2  —  5c 
-a  +  2  +    c 


8a 


.^G 


i^  —  hy 
y 


(8.) 
%x^  _  Sxy  +  2^*  +    c 
x^  —  %xy  +  oy'^  —  2c 
7a?'  +  Zxy  —   2/'  +  3c 

(10.) 
Sx^  —  Ixy  4-  21a  +    c 
—  x^  -\r  oxy  —   4a  +  4c 


(9.) 

ab  +  cd  —   m' 

ab  —  cd  —  2m^ 

2cd+    m' 

(11.) 
Sax  —     *iby  +  4:ab 
—  ax  — 10  by  -f  2a6 


12.  From  Sxy  —  20  subtract — xy  +  12.  Ans,  9xy  —  32. 

13.  From  7a^a;  +  a  subtract  3a^^  —  2a.  Ans.  ^a^x  +  3a. 

14.  From  —  8a;  —  2i/  +  3  subtract  lOa?—  Sy  -f  4. 

Ans.  —  18a?  •{■  y  —  1. 

15.  From  %y^  —  2y  —  5  subtract  —  Sif  —  5?/  +  12. 

Ans.  Uy^  +  Sy^n. 

16.  From  Tm^  — .  4a6  —  c  subtract  2m^  +  3c  —  8a&  —8. 

Ans,  5m'  +  4a6 — 4c  +  «, 
17    From  a  +  2a;  take  a  —  x.  Ans.  3^ 

4 


38  ENTIRE  QUANTITIES. 

18.  From  4a  -f  46  take  b  +a.  Ans.  3a  +  36. 

19.  From  4a  —  46  take  3a  +  56.  Ans.  a  —  96. 

20.  From  13a26'  +  11a  —  5a^  +  66,  take  7a  —  ^a"  +  66 
^  10a^6l  Ans,  2Za^W  +  4a. 

21.  From  3a  +  6  +  c  —  J— 10,  take  c  +  2a  —  d. 

u4ns.  a  +  6  —  10» 

22.  From3a  +  6+c  — cZ  — 10,  take  6  — 19  4- 3a- 

Ans,  c  — '  (Z  4-  9. 

23.  From  2a6  +  6^  —  4c  +  &c  —  6,  take  3a^  —  c  +  h\ 

Ans,  2ab  —  ^c  +  hc  —  3a'  —  b. 

24.  From  a'  4-  36'c  +  a6'  ■— a6c,  take  6'  +  a6'  —  a6c. 

Ans.  a^-\-Wc  —  h\ 

25.  From  bx^y  —  36a?  4-  e,  take  3a?V  +  26^27  -f  c^ 

Ans.  2x^1/  —  56^  —  d^  -{-c, 

26.  From  4m*  —  m  4-  2ca;  —  2/^  ^^^^  V^  —  ^^^ —  m  4-  car. 

Ans.  7m^  4-  ox  —  2y\ 

Note.  —  The  minus  sign  before  a  parenthesis  indicated  that  the  whole 
quantity  inclosed  is  to  be  subtracted. 

.      27.  What  is  the  value  of  3a'  —  (Sa  —  x  +  b)? 

Ans,  3a'  —  3a  4-  ^  —  6. 

28.  What  is  the  value  of  ^Oxy  —  (30^?/  —  26'  4-  3c  —  4:d)  ? 

An^s,  lOxy  +  26'  —  3c  4-  4 J. 

29.  What  is  the  value  of  a'  —  a  —  (4a  —  y  —  8a'  —  1)  f 

Ans.  4a'  —  5a  4-  2/  4- 1. 

80.  What  is  the  value  of  7m'  4-  26c  —  (3m'  _  6c  —  a?)  ? 

-4ns.  4m'  4-  36c  4-  x. 

81.  What  is  the  value  a4-^  —  ^  —  (wi  —  a  —  6)? 

Ans.  2a  4- 26  — 2m. 
54.  The  difference  between  two  dissimilar  quantities  may 
often  be  conveniently  expressed  in  a  single  term,  by  making 
some  common  letter  or  letters  the  unit  of  subtraction. 

Explain  the  unit  of  subtraction. 


SUBTRACTION.  g9 

1.  From  ax  subtract  hx. 

OPERATION.  .  Ti.  •         'A      LL\.   L-UL- 

Analysis.     It  is  evident  that  o  times 

Minuend,  ax  x  taken  from  a  times  x  must  leave  a 

Subtrahend,  bx  minus  b  times  X,  which  is  expressed 

7  rT~  thus  (a  —  b)x. 

Remainder,     {^a  —  u)X 

EXAMPLES  FOR  PRACTICE. 

(2.)  (3.)  (4.)  (5.) 

From                2am  my^  axy  ex 

Take                   cm  ny^  — cxy  x 

Rem.      (2a  —  c)m  (m  —  Ti)y^  (a  -\-  c)xy  (c  —  l)a7 

6.  From  2abx'^  take  bcx^,  Ans.  (2ab  —  bc)x\ 

7.  From  4:xy  take  mxz.  Ans,  (iy  —  mz)x. 

8.  From  ax  -i-bx  +  ex  take  x-h  ax  -{■  bx,    Ans,  (c  —  l)x. 

9.  From  3a^  —  by  take  2a^  —  cy,       Ans.  a"^  +  (c  —  b)y. 

10.  From  bacx^+20aa^y^-'2bm  take  3acx^+12ax^y^—207n, 

Ans.  2aa^(cx  +  4y')  -—  5m. 

11.  From  (2a  +  b  +  c)x  take  (a  +  b^x.     Ans,  (a  +  c)x, 

12.  From  (3a  +  c)xy  take  2axy  +  cxy.  -4ns.  axy, 

13.  From  at/  +  26i/  —  c^/  take  ai/  +  cy. 

'  Ans.  (2b  —  2c)y. 

14.  From  mz  take  n2; —  5z.  -4ns.  (m  —  n  +  5)z. 

15.  From  5a^x  —  2x  take  3a;  +  5aj?. 

^ns.  (5a' — 5a  —  5)a:. 

16.  From  —  ^G\m^  —  1)  take  7c'^(m'  —  1). 

Ans.  _10c=^(m'  — 1). 

17.  What  is  the  value  of  ^cy  —  x'^y —  {my  —  2x^y  +  2cy)  ? 

Ans.  (c  +  a;'  —  m)y, 

18.  What  is  the  value  of  3m —  z  —  y  —  (2z  —  y  —  3m)  ? 

!  Ans,  6m — ^z. 


40  ENTIRE  QUANTITIES. 


MULTIPLICATIOK 

i5fS.  Multiplication,  in  algebra,  is  the  process  of  taking  one 
quantity  as  many  times  as  there  are  units  in  another. 

CASE  I. 

56.  When  both  factors  are  monomials. 

1.  Multiply  4a  by  3&. 

OPERATION.  Analysis.   Since  it  is  immaterial  in  what 

A^        order  the  factors  are  taken  in  multiplica- 

Multiplicand,       fttt  ..  ,    ,,  _    ,.  '■. 

^y         tion,  we  may  proceed  thus  :  6'  times  4  are 

ipier,       -j^2.   ^  times  a  are  ah;   and  12  times  ab 

Product,  12ab     are  12a&,  the  entire  product. 

2.  Multiplya^  by  al 

OPERATION.  Analysis.     Since  a^  is  equal  to  aaa,  and 

•  a^   is  equal  to  aa.  their  product  must  bo 

Multiplicand,    a  ^  ,  .   ,      .  i    ,         *      .i  • 

2  aaaaa,   which   is  equal  to  a^;  this  expo- 

Multiplier,       a_  ^^^^^  5^  ^^y  ^^  ^^^^^  ^^  ^^^jjjg  ^j^Q  ^^^ 

Product,  a^  given  exponents,  3  and  2. 

•    3.  Multiply  3a6^  by  4^>». 

OPERATION. 

o^i^2         Analysis.    4  times  3a  are  12a ;  and  5^ 

^  ^^  ^      *     A   1.3     ti°^6s  ^^  ^re  6^ ;  hence  the  entire  product 
Multiplier,         4  6      jg  j2a  times  6«,  or  12a6«. 
Product,  12a6'^ 

Sy.  In  the  examples  given  above,  all  the  quantities  are 
understood  to  be  positive.  It  is  necessary,  however,  to  investi- 
gate the  law  of  signs  when  one  or  both  the  factors  are  negative 
In  arithmetic,  multiplication  is  restricted  to  the  simple  idea  of 

Define  Multiplication.    What  i|  Case  I? 


MULTIPLICATION.  41 

repeating  a  n amber.  In  algebra,  quantities  have  two  qualities, 
and  are  either  positive  or  negative  ;  and  multiplicaticn  has  the 
double  province  of  repeating  by  additions^  and  repeating  by 
subtractions,  as  indicated  by  the  signs  of  the  multiplier. 
Hence,  the  full  signification  of  a  multiplier,  when  analyzed,  is 
as  follows  : 

I.  The  plus  sign  of  a  multiplier  shows  that  the  multiplicand 
is  to  be  added  to  zero. 

II.  The  minus  sign  of  a  multiplier  shows  that  the  multi- 
plicand is  to  be  subtracted  from  zero  ;  and 

III.  The  value  of  the  multiplier  sliows  how  many  times  the 
multiplicand  is  to  he  taken  by  either  process. 

To  exhibit  the  law  which  governs  the  sign  of  a  product, 
according  to  these  principles,  we  present  four  examples,  as 
follows : 

1.  Multiply  a  by  h. 

OPERATION. 

,  Analysis.     The  plus  sign  of  the  multiplier  indi- 

cates that  the  multiplicand,  +  a,  is  to  be  added  to 
zero     h    times,    giving  +  a  +  a  -f  a,    &c. ;    hence 


h 


"f  ^^         the  result  will  be  positive,  or  +  ah. 

2.  Multiply  —  a  by  —  5. 

OPERATION.  Analysis.     The  minus  sign  of  the  multiplier  in- 

—  a  dicates  that  the  multiplicand,  —  a,  is  to  be  subtracted 

—  h  from    zero    b   times,   which  will    change  its  sign, 
-\-  ah  g^'^^^S  -{•  a  -\-  a -{-  a,  &c. ;  hence  the  result  will  bo 

positive^  OT  -\-  ah, 

3.  Multiply  a  by  —  6. 

OPERATION.  Analysis.     The  minus  sign  of  the  multiplier  in- 

+     a  dicates  that  the  multiplicand,  +  a,*is  to  be  subtracted 

—  h  from    zero    h    times,    which   will    change  its  sign, 
'  "^  giving  —  a  —  a  —  a,  &c. ;  hence,  the  result  will  be 

negative,  or  —  ah. 

Explain  the  difference  between  Arithmetical  and  Algebraic  multiplica- 
tion. How  do  the  signs  -\.  and  —  before  a  multiplier  affect  the  product? 
The  value  of  a  multiplier  ehows  what  ? 

4* 


42  ENTIRE  QUANTITIES. 

4.  Multiply  —  ahj  b. 

OPERATION.        Analysis.    The  plus  sign  of  the  multiplier  indi- 

—    a         cates  that  the  multiplicand,  — a,  is  to  be  added  to 

+    5         zero      b    times,  which    only    repeats    the     letter, 

7T         giving  —  a  —  a  —  a,  &c. ;  hence  the  result  will  bo 

negative^  or  —  db. 

Comparing  the  four  examples,  we  observe  that  like  signs 
produce  plus  ;  and  unlike^  minus. 

From  the  foregoing  examples  and  illustrations  we  derive 
the  following 

Rule.  I.  Multiply  the  coefficients  of  the  two  terms 
together  for  the  coefficient  of  the  product. 

II.  Write  all  the  letters  of  both  terms  for  the  literal  part, 
giving  each  letter  an  exponent  equal  to  the  sum  of  its  expo- 
nents in  the  two  terms. 

III.  If  the  signs  of  the  two  terms  are  alike,  make  the  prO" 
duct  plus  ;  if  unlike,  make  it  minus. 

Note  1,  The  value  of  the  product  will  be  the  same  in  whatever  order 
the  factors  are  written.  Most  algebraists  prefer  to  arrange  letters  in 
alphabetical  order. 

EXAMPLES  FOR  PRACTICE. 

5.  Multiply  8^  by  7a.  Ans,  21ax, 

6.  Multiply  4y  by  Sab.  Ans.  12ab7j. 

7.  Multiply  156c  by  lOa?.  Ans.  IbObcx. 

8.  Multiply  Qax  by  12by.  Ans.  72abxy. 

9.  Multiply  17cd  by  3m.  Ans.  blcdm. 

10.  Multiply  4pg  by  7xy.  Ans.  2Spqxy. 

11.  Multiply  12am  by  ^bcd.  Ans.  60abcdm. 

12.  Multiply  2bpqr  by  Sxyz.  Ans.  Ibpqrxyz. 

13.  What  is  the  product  of  a^  by  a^  ?  Ans.  a\ 

14.  What  is  the  product  of  x^hj  x^?  Ans,  x^°. 

15.  What  is  the  product  of  y^hj  y^l  Ans.  y^^. 

If  the  factors  have  like  signs,  what  must  be  the  sign  of  the  product  ? 
If  unlike,  what  ?     Give  the  rule  for  multiplication  of  monomials. 


MULTIPLICATION.  48 

16.  What  is  the  product  of  m^  by  m^  ?  Ans.  m}\ 

17.  What  is  the  product  of  b^a^  by  6V  ?  ^ns.  6V. 

18.  What  is  the  product  of  aW  by  am^  ?        J^ns.  a^m^ 

19.  Multiply  4ac  by  — .3a6.  -4ns.  —12  a'6c?. 

20.  Multiply  9a^G  by  —  4ay.  -4ns.  —  36a^c2/. 

21.  Multiply  —  2xy  by  —  2xy.  Ans.  4iX^y\ 

22.  Multiply  —  lay  by  3^?/.  .4ns.  —  21axy\ 

23.  Multiply  21^;^  by  —Zxy.  Ans.  —  63^2/'- 

24.  Multiply  —  5a^m  by  —  4:abm\  Ans.  20a^bm^. 

25  Multiply  —  Tm^  by  lOc^m'a:.  Ans.  —  lOc^m^z. 

26  Multiply  llxy  by  2a;'2/'-  ^^s.  34a75t/». 

27.  Multiply  Uab'cd^  by  —  363c2m. 

^ns.  — 42a6'^c'<:?^m. 

28.  What  is  the  value  of  3a  X  46  x  2c  ?        Ans.  24:abc. 

29.  What  is  the  value  of  7m^  X  4am  x  2my  ? 

Ans.  56am*i/, 

30.  What  is  the  value  of  ar*  x  a;'  X  ^  ?  Ans,  of, 

31.  What  is  the  value  of  —  la'b  X  2ab''  X  3a6  ? 

Ans.  —  42a*&*. 

32.  What  is  the  value  of  —  ba'm  X  3a6'c  X  2bc^m^  ? 

Ans.  — 30a''6Vm^ 
83.  Multiply  3(a;  +  2/)  ^7  2.  Ans.  6(a?  +  y'). 

34.  Multiply  a(^x^  +  m)  by  b,  Ans.  ab(x^  +  m). 

35.  Multiply  (a  -f-  wi)^  by  c.  ^ns.  c(a  +  w?,)\ 

36.  Multiply  (a  +  by  by  (a  +  b)\  Ans.  (a  +  6)\ 

37.  Multiply  3a(m  —  n)^  by  —  a(m  —  n)l 

Ans.  — 3a*(m  —  n)*. 

38.  Multiply  4m(j7^  —  y^)'^  by  —  2am(^=^  —  2/0- 

Ans.  —SamXx^  —  y^y. 

Note. — "When  quantities  have  literal  exponents,  powers  of  the  same 
letter  or  quantity  are  multiplied  by  indicating  the  addition  of  the 
exponents. 

39.  Multiply  a"»  by  a".  Ans.  a"'  +  ". 
40    Multiply  c"*  by  c.  Ans.  c"*+*. 

How  are  quantities  with  literal  exponents  multiplied? 


44  ENTIRE  QUANTITIES. 

41.  Multiply  (a  —  by  by  (a  —  by. 

Ans,  (>i  — 6)'+^ 
42    Multiply  a'"(p  +  qy  by  a'O  +  (?)"». 

CASE   II. 

58.  When  one  factor  is  a  polynomial. 

1.  Multiply  46  +  5a'  —  be  by  3a. 

Analysis.     Since  the  whole  multi- 

OPERATION.  plicand  is  to  be  taken  3a  times,  we 

.,    ,      p^  2       z,  must  multiply  each  of  its   terms  by 

4^  -^     Da  —  ^C  3^ .  ^j^^g  3^  ^jj^gg  4^  jg  ^2a6 ;  3a  times 

*^^ Sa'^  is  15^3 ;  3a  times  —  &c  is  —  3a6c ; 

12a6  +  15a'  —  Sabc         ai^d  "we  have  for  the  entire   product, 
12a6  +  I5a^  —  ^abc.    Hence  the 

Rule.  IluUiply  each  term  of  the  polynomial  separately 
by  the  multiplier j  and  write  the  partial  products  connected  bv 
their  proper  signs. 


(2.) 

5a  — 8c 
2a 

EXAMPLES  FOB.  PRACTICE. 

(3.)                                  (4.) 
3ac  — 46                  2a'— 8c +  5 
—  3a                             he 

10a'  — 6ac 

(5.) 
12j;  — 2ae 

4a 

—  Wc  +  12a6 

(6.) 
15c— 76 
—  2a 

2a'6c  — 36c''  +  56c 

(7.) 

4a;  — 6  +  3a& 
2a6 

(8.) 

8c'  4-  X 
\xy 

(9.) 
—  4x' 

(10.) 
3a'  — 2a;' —  66 

2aa;' 

Give  Case  II.     Analysis.     Rule. 


MULTIPLICATION.  45 

11.  Multiply  36  —  2c  by  56(?  Ans.  l^b'^c  —  I0bc\ 

12.  Multiply  ^xy  —  9  by  6x,  Ans.  2Aa;^y  —  54^7. 

13.  Multiply  a^  —  2^  +  1  by  4:x\ 

-       Ans,  4:aV---Sx'  +  4:x' 

14.  Multiply  lla^6c^  —  ISxy  by  Sax, 

Ans.  SSa'^bc^x  —  S9ax% 

15.  Multiply  42c'  —  1  by  —  4.  Ans.  —  168c'  +  4. 

16.  Multiply  —  30a'6a;V  +  13  by  —  oa\ 

Ans,  l^Oa^bx^y  —  65a' 

17.  Multiply  26  —  7a  —  3  by  4a6. 

Ans,  8a6' —  28a'6  —  12a6. 

18.  Multiply  a  +  36 —  2c  by  —  3a6. 

Ans.  —  3a'6  —  9a6'  +  6a6c. 

19.  Multiply  13a'  —  6'c  by  —  4c. 

Ans,  —  52a'c  +  46V. 

20.  Multiply  ISxy  —  36  by  —  25^. 

^  Ans,  —  S2bx^y  +  756a?'. 

CASE  in. 

59.  Wbieii  both  factors  are  polynomials. 

1.  Multiply  2a  +  36  by  a  +  6. 

OPERATION.  Analysis.   To  multiply  by 

Multiplicand,  2a    -r  36  a+  6,  we  must  take  the  mul- 

Muitipiier,  a    +     6  tipKcand  a  times  and  6  times, 

Product  by  a,  2a''  +  3a6  ^hich  is  done  by  multiplying 

Product  by  6,  +  2a6  +  36'  ^y  f  ^^^  ^  separately,  and 

„  ..    ^    ,    .  o  ■>    ■ — r    7    .    or9  adding  the  partial  products. 

EnUre  Product,  2a^  +   5a6  +  36'  -rj           th 

Rule.  Multiply  all  the  terms  of  the  multiplicand  by  each 
ierm  of  the  multiplier  separately ^  and  add  the  partial 
products. 

Give  Case  III.    Analysis.    Rule. 


46  ENTIRE  QUANTITIES. 

EXAMPLES  FOR  PRACJTICB. 


Multiply 

(2.) 
2a   +  6c 

(3.) 
8^  —  5y 

a  —   c 

X  —2y 

Product, 

2a'  4-  5ac 

_2ac  — 5c* 
2a' 4-  2>ac  —  h(^ 

Bx^  —  bxy 

—   Qxy  +  lOy" 
Zx'  —  llxy+lOy' 

Multiply 

By 

Product, 

(4.) 
^x  —2z 
3a   —bd 
ISax  —  Qaz — SOdx  +  lOdz 

Multiply 
By 

(5.) 
a-h    b  +    c 

X  +    y  +    z 

Product, 

ax-i-  bx  +  ex  +  ay+  by  +  cy  +  az  +  bz  +  cz. 

6.  Mult 

iply  Sa^  —  2a6  —  b^  by  2a  - 

-4b, 

Ans,  6a»  —  IQa'b  +'^ab'  +  W. 

7.  Multiply  x'^  —  xy  +  y^hj  X  +  y.  Ans.  a^  +  y^. 

8.  Multiply  3a  4-  4c  by  2a  -—  5c. 

Ans.  6a'— 7ac  — 20c*. 

9.  Multiply  a^  +  ay  —  y^hy  a  —  y. 

Ans,  a*  —  2ay^  +  y^. 
10.   Multiply  a^  +  ay  +  y'^hj  a  —  y,  Ans.  a"  —  y^. 

p.1.  Multiply  a'  —  ay  +  y^hj  a  +  y.  Ans.  a'  4- 1/*. 

12.   Multiply  a^  +  a^y  +  ay^  +  y^hj  a  —  y. 

Ans.  a*  —  y*. 

13  Multiply  y^  —  y  +  lhj  y  +  1,  Ans.  t/*  4- 1. 

14  Multiply  a^  +  y'^hj  x^  —  i/l  Ans.  a^  —  t/*. 

15.  Multiply  a*  —  3a  4-  8  by  a  4-  3.      Ans.  a»  —  a  4-  24. 

16.  Multiply  b"  4- 1 V  4-  a;*  by  6'  —  x\        Ans.  5« — 3^. 

17.  Multiply  a*  4-  26  by  2a^  —  46.  Ans.  2a*  —  86*. 


MULTIPLICATION.  47 

18    Multiply  a^  -]'  x^  +  x^  hj  x^  —  1.  Ana,  x^  —  x\ 

19.  Multiply  m  +  n  by  9m  —  9n.  Ans.  9m^ — dn\ 

20.  Multiply  2x^  +  xy  ^  "If  by  Zx  —  3?/. 

Ans.  Ga?'  —  Zx^y  — ^xy^  +  63/*. 

21.  Multiply  m'  —  3m  —  7  by  m  —  2. 

Ans.  m^  —  5m^  —  m  +  14. 

22.  Multiply  a*  —  2a«c  +  4a'c'—  8ac^  +  16c*  by  a  +  2e. 

23.  Multiply  a;^  —  3^?^  +  3a;  —  9  by  a?  +  3. 

^ns.  a;*  — 6ar^  — 27. 

24.  Multiply  m*  —  m*  +  m*  —  m  +  1  by  m  +  1. 

Ans.  m^  +  1. 

25.  Multiply  m*  +  w<'  +  w'  +  ^  +  1  by  m  — 1. 

26.  Multiply  2a3 '+  5ac'  —  2c^  by  2a»  —  5ac'  +  2c'. 

^ns.  4a«  —  25aV  +  20ac*  —  4c^ 

Note.  —  The  product  of  two  or  more  polynomials  may  be  indicated  by 
inclosing  each  in  a  parenthesis,  and  writing  them  in  succession ;  siich  an 
expression  is  said  to  be  expanded,  when  the  multiplication  has  been 
actually  performed. 

27.  Expand  (a  +  6)  (a  +  c). 

Ans.  a?  +  ah  +  ac  +  be. 

28.  Expand  (x  +  By)  (x^  —  2/)- 

Ans.  xr^  +  Bx^y  —  xy  —  By\ 

29.  Expand  (w}  +  2c)  (m^  —  5c). 

80.  Expand  (a  +  6  —  c)  (a  —  6  +  c). 

Ans.  a'  —  l)'  +  2hc  —  &. 
31.  Expand  (a  —  c  —  1)  (a  +  1). 

Ans.  d^  —  ac  —  c  —  1. 

CASE   IV. 

60.  To  square  a  binomial. 

If  a  polynomial  be  multiplied  by  itself,  the  product  is  the 
square   of  the  polynomial.      A  binomial  quantity  is  easily 

How  may  multiplication  of  polynomials  be  indicated?  When  are  such 
expressions  expanded  ?    Give  Case  IV. 


48  ENTIRE  QUANTITIES. 

squared  without  the  formal  process  of  multiplying,  as  will  be 
seen  by  the  analyses  of  the  two  following  examples. 

1,  What  is  the  square  of  a  +  6  ? 

OPERATION. 

Analysis.    Multiplying  a  -|-  6  by 

'  a  -j-  6  by  the  common  method,  wo 

(i  ~\-  0  obtain  for  the  result,  a^  which  is  the 

^2  _j_  ^jy  square   of  a;    -j-  2a6,  which  is   the 

-A-  ah    4-  h^  ^  times  product  of  a  and  6 ;  and  6^, 

■  which  is  the  square  of  6. 

052  _|.  2ah  +  6^ 

2.  What  is  the  square  of  fl  —  6  ? 

OPERATION. 


a  — b 

a  —  h 


Analysis.  Multi  plying  in  the  usual 
way,  we  obtain  for  the  result  a^  which 
is  the  square  of  a ;  —  2a5,  which  is 
a^  —  ah  twice  the  product  of  a  and  —  h ;  and 
dj)    +6^  ^^  which  is  the  square  of  6.     Hence 

o?  —  2ah+h'  *^® 

Rule.     Write  the  square  of  the  first  term^  twice  the  pi\ 
duct  of  the  two  terms,  and  the  square  of  the  second  ter^m. 

Note. — The  product  of  the  two  terms  will  be  minus,  when  one  of  them 
Is  minus. 

EXAMPLES  EOR  PRACTICE. 

3.  Square  a  +  c.  Ans,  d^  +  2ao  +  &. 

4.  SquaBe  p  +  ^-  Ans,  p^  +  2pq  +  (f 

5.  Square  m  —  n.  Ans,  m?  —  2mn  +  ^^ 

6.  Square  x  —  y,  Ans,  x^  —  2xy  +  2/' 
7-  Square  A  +  B.  An^.  A^  +  2AB  +  B 

8.  Square  A—G.  Ans,  A'  —  2AG+  G\ 

9.  Square  8a  — 2^.  Ans,  Oa^  —  12aa7  +  4ar*. 


Give  analysis     Rule. 


MULTIPLICATION.  49 

10.  Expand  (m  +  z)  (m  +  z).        Ans.  m^  +  2mz  +  z\ 

11.  Expand  (2a  —  c)  (2a  —  c).      Ans.  4a^  —  4ac  +  c^. 

12.  Expand  (5^7  —  3)  (5a;— 3).      ^ns.  25a;2_3o^  +  9 

13.  Expand  (4a  +  i^)  (4a  +  ^x), 

Ans,  16a^  +  4:00)  +  ^a;'. 
Note. — The  square  of  a  binomial  may  be  indicated  by  an  exponent. 

14.  Expand  (m  +  cy,  Ans,  m^  +  2cm  +  c\ 

15.  Expand  (2c  —  Sdy,  Ans,  4c^  _  12cd  +  9cP. 

16.  Expand  (x^  —  xy,  Ans,  a^--2cc^  +  x\ 

17.  Expand  (a  — 1)1  Ans,  a'  —  2a  +  1. 

18.  Expand  {a^x  —  ax^y,  Ans.  aV  —  2aV  +  aV. 

19.  Expand  {y^  —  20)  {y''  —  20)> 

^ns.  2/*  —  402/' +  400. 

20.  Expand  (a:"*  —  y")  {x"^  —  2/"). 

^ns.  a;''"  —  2^?"'?/"  +  2/'".      - 

21.  Expand  (c^  •—  1)  (c'"  —  1).       ^ns.  c'"*  —  2c^  +  1. 

CASE   V. 

01.  To  find  the  product  of  the  sum  and  difference 
of  two  quantities. 

The  sum  of  two  qnantities  multiphed  by  their  difference 
gives  a  result  still  more  simple  than  a  binomial  square. 

1.  Multiply  a  +  &  hy  a  —  h. 

OPERATION.  Analysis.     We  are    required  to 

Sum,  a  -\-  h  multiply  the  sum  of  a  and  h,  by  the 

,  difference  of  a  and  b.     Multiply  in  a: 

Difference,    a   0  ,  •^\,  ,  „     ,  .  ' 

by  the  usual  process,  we  find  in  add- 

a^  +  <^^  ing  the  partial  products,  that  -f  ab 

. —  ab  —  b^  and  —  ab  reduce'  to  zero,  and  the 

— ^ — product  is  a^  —  6^  or  the  difference 

Product,     a         0  of  the  squares  of  a  VLud  b.     But  since 

«  and  b  may  represent  any  two  quantities  whatever,  the  form  of  this 
product  embodies  a  general  truth.     Hence 

Give  Case  V.     Analysis. 
5  D 


50  ENTIRE  QUANTITIES. 

Rule.     From  the  square  of  the  greater  quantity ^  subtract 
the  square  of  the  less. 

NoTB. — The  term  or  quantity  having  the  minus  sign  in  the  difference, 
is  supposed  to  be  the  less. 

EXAMPLES  FOR  PRACTICE. 

2.  What  is  the  product  of  m  +  n  by  m  —  n? 

Ans,  w?  —  n* 
8.  What  is  the  product  of  a  +  c  by  a  —  c? 

Arts,  a^  —  <f. 

4.  What  is  the  product  of  A  +  Bhj  A  —  B? 

Ans,  A'  —  B". 

5.  What  is  the  product  of  2m  +  2n  by  2m  —  2n? 

Ans.  4:m^  —  4n*. 

6.  What  is  the  product  of  x  +  yhj  x  —  y? 

Ans.  x^  —  2/'- 

7.  What  is  the  product  of  Sx  +  Sy  by  Sx  —  Sy? 

Ans.  9x^  —  9y\  • 

8.  What  is  the  product  of  7a  +  6  by  7a  —  6  ? 

Ans.  4:9a''  —  b\ 

9.  What  is  the  product  of  1  +  10a  by  1  —  10a? 

Ans.  1  —  lOOal 

10.  Expand  (1  — c")  (1  +  c"*).  Ans.  1  —  c^"". 

11.  Expand  (1  +  2^)  (1  _  2x).  Ans.  1  —  ^x". 
—'12.  Expand  (a  +  ^x)  (a  —  \x),                Ans,  a^  —  Ja?l 

«^13.  Expand  (x  +  y  +  z)  {x  +  y  —  z). 

Ans.  (x  +  yy  —  z\ 
^•^14,  Expand  (1  +  m — c)  (1  —  m  —  c). 

Ans.  1  —  (m  —  cy. 

^  15.  Multiply  (a  +  6)  +  (a?  +  2/)  by  (a  +  6)  —  (a;  +  2/) 

Ans.  {a-\'hy  —  {x+yy. 

Note.  —  To  simplify  example  15,  let  P  =  a  +  5,  and  Q  =  a;  +  y,  then 
the  product  required  will  be  (P  4.  Q)  (P  —  Q)  =  pa  —  Q«. 

Give  Rule. 


MULTIPLICATION. 


51 


^  16.  What  is  the  product  of  a  +  'Ixy  x  a  —  2xy  x  56  ? 

A71S.  5a'b  —  20bxy, 
-^17.  What   is   the    product    of   (2^^*  +  3i?"a^)  (2^1-  — 

O^.  By  applying  the  principles  established  in  the  last  two 
articles,  much  labor  may  often  be  saved  when  it  is  required  to 
find  the  product  of  three  or  more  binomials. 


1.  What  is  the  value  of  (x  +  3)  (x  +  3)  (x  -f  3)  whcD 
expanded  ? 


OPERATION. 

x'  +  6x  -f  9 
X  +S 

x^  +  Qx'+    9x 

Sx'  4-  18^  +  27 


ptoduct,  a^  +  9.r^  4-  2707  +  27 


Analysis.  We  are  able  r-t 
once  to  VTi'iie  the  product  of 
the  first  two  factors,  (x  -{-  Z) 
{x-\-  3),  which  isa;*4-  6.%'  -j-  9, 
(60) ;  and  multiplying  by 
a?  4-  3,  the  other  factor,  we  ob- 
tain ar»  +  9a;2  +  27a:  -f  27,  the 
final  product. 


2.  Expand  (a  -{•  b)  (a  —  6)  (a  —  c). 


OPERATION. 

(a  +  6)(a  — 6)  =  a^  — 5'' 
a  —  c 


Analysis.    We  ex- 
pand by  the  rule  (61), 
(a  -f  h)  (a  —  b)y  and 
then  multiply  the  re- 


Product,      a^  —  ab^  —  a^c  +  b^c    suit  by  a  —  c. 


8.  Expand,  (x  —  c)(x  —  d)  {x  -}-  c)  {x  +  d). 


OPERATION. 


Analysis.  We  write  x^  — fi^, 
the  product  of  the  first  and 
third  factors ;  and  under  this 
x^  —  d\  the  product  of  the 
second  and  fourth  factors ; 
then  we  multiply  these  two 


products  together,  and  obtain  the  product  of  the  four  factors. 

■■-.-a.. 


V^T. 


■^ 


^ 


62  ENTIRE  QUANTITIES. 

Note.  —  In  obtaining  the  continued  product  of  several  factors,  tht 
pi'pil  should  use  judgment  in  selecting  and  combining  the  factors,  so  aa 
to  enable  him  as  far  as  possible  to  write  out  the  results  mentally. 


EXAMPLES   FOR  PRACTICE. 

4.  Expand  c(m  —  n)  (m  +  n).  Arts,  cm*  —  cn\ 

6.  Expand  (3a  —  b)  (3a — b)x, 

Ans.  da^x  —  Qabx  +  6^^« 

6.  Expand  (2m  —  c)  (2m  -f  c)  (4m'  +  c'). 

Ans.  16m*  —  <7*. 

7.  Expand  (a  +  c)  (a  +  cf)  (a  —  c)  (a  — J). 

Ans.  a* —  aV  —  a'd'  +  c^(P. 

8.  Expand  (1  +  c)  (1  -f  c)  (1  —  c)  (1  +  c^. 

^ns.  1  +  c  —  c*  —  c*. 

9.  Expand  (^  —  4)  (^—5)  (x  +  4)  (^  +  5). 

Ans.  x'  —  41^-2  -f  400. 
—  10.  Expand  (Sx  —  ??i)  (x'  +  m')  (3^  —  7?i). 

Ans.  9x*  —  C^x^m  -f  1  Ox'-^m'  —  6xm'  +  m*. 
il.  Expand  (2a  +  ox)  {'la  -j-  3./;)9. 

^?is.  36a'  +  108a:c  +  Ux\ 

12.  Expand  (led'  -f  42/^2:)  (led'  —  iyh'). 

Ans.  ^9c'd'  —  16y'z\ 

13.  Expand  (x  +  1)  (^  +  1)  (^x  —  2). 

Ans.  a^  —  Sx — 2. 

14.  Expand  (m  —  2)  (m  —  2)  (m  +  1). 

Ans.  m'  —  3m'  +  4. 

15.  Expand  (m}^  +  1)  (m»  +  1)  (m*  +  1)  (m*  +  1)  (m  +  1) 
(m —  1).  Ans.  m"'*  —  1. 


DIVISION. 


DIVISION. 

63.  Division,  in  Algebra,  is  the  process  of  finding  how 
many  times  one  quantity  is  contained  in  another.  It  is  the 
converse  of  multiplication,  the  dividend  answeriug  to  the  pro- 
duet,  and  the  divisor  and  quotient  to  the  multiplier  and  midU 
tiplieand. 

CASE   I. 

64.  When  botli  dividend  and  divisor  are  monomials. 

1.  Divide  Qab  by  2a. 

OPERATION.  Analysis.    Since  division 

n   1  _j_  9     Of  is  the  converse  of  multipli- 

cation, we  must  seek  for  a 
quantity,  which  mj.dplied  by  2a,  the  divisor,  will  produce  Ca6,  the 
dividend.  This  quantity  is  36,  and  is  found  by  inspection^  or  by 
dividing  6,  the  coefficient  of  the  dividend,  by  2,  the  coefficient  of  the 
divisor,  and  dropping  the  factor  a,  common  to  both  dividend  and 
divisor. 

2.  Divide  ^  by  x}. 

OPERATION.  Analysis.    Since  the  divi- 

a^  -^  of  =  a;*---  "^^^  multiplied  by  the   quo- 

tient will  produce  the  divi- 
dend, the  exponent  of  the  quotient  must  be  such  as,  when  added  to  the 
exponent  of  the  divisor,  will  be  equal  to  that  of  the  dividend.  Hence, 
we  subtract  2,  the  exponent  of  the  divisor,  from  5,  the  exponent  of 
the  dividend,  and  obtain  3,  the  exponent  of  the  quotient. 

Note.  —  To  midtiphj  one  power  by  another  of  the  same  letter,  we  add 
exponents ;  to  divide  one  power  by  another  of  the  same  letter,  we  sub- 
tract -exponents. 

3.  Divide  a^mv'z^  by  771^2;*. 

OPERATION.  Analysis.     The  exponent 

a^iiv'z^  -^  ni^:^  z=  a^vi^  ^^  ^  ^^  *^^  quotient  is  2  ;  the 

exponent  of  w  is  5  —  2  =  ii ; 
and  the  exponent  of  2;  is  3  —  3=0;  this  signifies  that  z  is  taken  no 
times  in  the  quotient,  and  is  therefore  cancelled. 

Define  Division.     Show  its  relation  to  Multiplication.  >^ 

5* 


'!f^  ENTIRE  QUANTITIES. 

05,  In  the  foregoing  examples,  no  signs  being  expressed, 
both  dividend  and  divisor  are  understood  to  be  positive.  To 
ascertain  what  sign  the  quotient  should  have  when  one  or  both 
the  terms  of  division  have  the  minus  sign,  we  have  only  to 
observe  what  sign  must  be  given  to  the  quotient,  in  order  that 
the  product  of  the  quotient  and  divisor  shall  have  the  same 
sign  as  the  dividend,  according  to  the  law  of  signs  in  multi- 
plication. 

To  exhibit  the  law  which  governs  the  sign  cf  the  quotiei^t, 
Fe  present  four  examples. 

1.  +ab-7-  +  a=  +  b  because  -{-ax  -{-  b  =  -\-  ab. 

2.  —  ab  -7--^a=  +  b        ''  —  a  x  +  b  =  —  ab. 

*S,   +  ab-. a  =  —  b        "  —  ax  —  b  =  +  ab, 

4.  —  a5  _i_  _|_  a  =  —  b        "  +  a  X  —  b  =  —ab. 

From  these  examples,  taken  in  order,  we  make  the  following 
inferences : 

IsL   4-  divided  by  +  e-ives  +7         n       •  ^  ^     i 

^  .7    J    b  •>-  or,  like  signs  produce  -|- 

2d.    —  divided  by  —  gives  +  ) 

6d.    +  divided  by  —  gives  — 7  n       •  i 

,.  .,    ,  ^  .  >■  or,  unhke  signs  produce  — 

it/L  —  divided  by  +  gives  — ) 

G©.  These  priuciples  may  be  deduced  from  tlie  nature  of 
the  signs  themselves,  by  taking  another  view  of  division. 

Division,  considered  in  its  most  elementary  sense,  is  not 
merely  the  converse  of  multiplication;  it  is  a  short  process 
of  finding  how  many  times  one  quantity  can  be  subtracted  from 
ciuothcr  of  the  same  kind.  When  the  subtraction  is  possible, 
and  diminishes  the  numeval  value  of  the  minuend,  and  brings 
it  nearer  to  zero,  the  operation  is  real  and  must  be  marked 
plus.^  When  the  subtraction  is  not  possible  without  going 
farther  from  zero,  we  must  take  the  converse  operation,  and 
the  converse  operation  we  must  mark  min^is. 

Give  analyses  of  the  law  which  governs  the  signs  of  the  quotient.  Give 
the  law- 


DIVISION.  55 

Thus,  divide  18a  by  Qa.  ITere,  it  is  proposed  to  find  how 
many  times  6a  can  be  subtracted  from  18a ;  and  as  wo  can 
actually  subtract  it  3  times,  the  quotient  must  be  -f  3- 

Divide  —  18a  by  —  im.  Here,  again,  the  subtraction  can 
actually  he  performed,  and  the  number  of  times  is  o,  and,  of 
coarse  the  quotient  is  -f-  '^• 

Divide  —  I8a  by  Oa,  Here,  subtraction  will  not  I'educe  the 
dividend  to  zero ;  but  addition  will,  and  must  be  performed  3 
times ;  but  the  operation  is  the  converse  of  the  one'proposed, 
and  therefore  must  be  marked  by  the  converse  sign  to  pluSy 
that  is  —  3. 

Again,  divide  18a  by  —  Co.  Here,  if  we  subtract  —  6a  it 
will  not  reduce  18a ;  but  the  converse  opemtion  will,  and 
therefore  the  quotient  must  be  minus,  that  is,  —  3. 

From  all  these  illustrations  we  derive  the  following : 

Rule.  '  I.  Divide  the  coefficient  of  the  dividend  by  the 
coefficient  of  the  dimsor^  for  a  new  coefficient. 

II.  Write  the  lette^^s  of  the  dividend  in  the  quotient ^  giving 
each  an  exponent  equal  to  tlie  difference  of  its  exponents  in 
the  two  terniSj  and  suppressing  all  letters  whose  exponents  be- 
come zero. 

III.  If  the  signs  of  the  terms  are  alike ^  make  the  quotifi*'t 
plus  ;  if  unlike,  make  it  minus. 

Note. — If  the  dividend  does  not  exactly  contain  the  divisor,  the  division 

may  be  indicated  by  writing  the  dividend  above  a  horizontal  line,  and 

the  divisor  bo1ow,  in  the  form  of  a  fraction ;  and  the  result  thus  obtained 

may  be  simplified  by  canceling  all  the  facfbrs  common  to  the  two  terms ; 

ia^b'^c        2b 
thus,  ^a'^b^c  ~  Qii'^b^c"^  :=  rr:-— .     But   this   process   is  essentially 

a  case  of  reduction  of  fractions;  we  shall  therefore  omit  all  examples  of 
this  class  till  the  section  on  fractions  is  reached. 

EXAMPLES   fOR   PRACTICE. 

1    Divide  16a&  by  4a.  Ans.  4&. 

2.  Divide  21acd  by  7c.  Ans.  Sad. 

3.  Divide  ab'c  by  ac.  Ans.  b\ 

Kule  for  division  of  monomials. 


56  ENTIRE  QUANTITIES. 

4.  Divide  6abc  by  2c.  Ans.  SaO. 

5.  Divide  aoc^  bj  ax\  Ans.  x, 

6.  Divide  Bmx^  by  mx.  Ans,  ^x^, 

7.  Divide  210c'b  by  7c6.  u4ns.  30cl 

8.  Divide  42a;?/  by  xy,  Ans.  42. 

9.  Divide  —  2 lac  by  —  7a.  Ans.  3c. 
10  Divide  —  12xy  by  By.  -4ns.  —  4^7. 

11.  Divide  72a&c  by  —  8c.  Ans,  —  9ab 

12.  Divide  2a«  by  a*.  Ans,  2d\ 

13.  Divide  —  a^  by  a*^.  u^ns.  —  a. 

14.  Divide  16^  by  4aj.  ^ns.  4j;^ 

15.  Divide  Ibaxy^  by  —  oa^/.  u4ns.  —  bxy'^, 

16.  Divide  *—  ISa^x  by  —  6a^.  u4ns.  oa\ 

17.  Divide   Qacdxy^  by  2adxy^  Ans,  3c. 

18.  Divide  12aV  by  —  3a^^.  ^ns.  —  4x. 

19.  Divide  15ai/^  by  —  Say.  Ans,  —  by. 

20.  Divide  45(a  t-  xf  by  15(«  —  ^)^  Ans.  3(a  —  a;). 
Note. — In  this  example,  copslder  (a  —  z)  as  one  quantity. 

21.  Divide  45i/^  by  15y^.  ^ns.  3^/. 

Note.  — Examples  20  and  21  are  exactly  alike,  if  we  conceive  (a  —  x) 
equal  to  y. 

22.  Divide  z^  by  zK  Ans.  z\ 

23.  Divide  (x  —  ^z)^  ^7  (^  —  2/)'-  -^^s*  (^  —  2/)^- 
Note.  —  Observe  that  examples  22  and  23  are  essentially  alike. 

24.  Divide  (a  +  b'^^hj  (a  +  6).  Ans.  (a  +  6)1 

25.  Divide  x"^  by  a;^  -4ns.  ^'""■", 

26.  Divide  6c^  by  3c.  Ans.  2c'^-'K 

27.  Divide  (a  —  c)"'  by  (a  •—  c)l  ^ns.   (a  —  c)'"--^ 

28.  Divide  10(a  —  c)  by  5(a  —  c).  Ans.  2. 

29.  Divide  .6a%a  +  my  by  ^a  +  m) 

-4ns.  S^fa  +  m)^ 

30.  Divide  52m^c(l  —  x'y  by  13mc(l  —  x'f, 

Ans.  4m. 

31.  Divide  81a*;s'^(4m  —  qf  by  27 zX^m  —  qy 

Ans.  3a*(4m  —  q). 


division;.  57 

CASE   II. 

67.  To  divide  a  polynomial  by  a  single  term. 

I.  Divide  I2a'  —  Qa'c  +  Ba'm  by  3al 

OPERATION.  Analysis.     The  whole   divi- 

Sa^)12a^  —  Qa^c  +  3a^m  dend  is  divided  by  'Sa\  by  di- 

4^3 2ac  4-  m  viding  each  of  its  terms  by  Za\ 

Heuce  the 

Rule.     Divide  each  term  of  the  dividend  separately ^  and 
connect  the  quotients  by  their  proper  signs. 

EXAMPLES   FOR   PRACTICE 

2.  Divide  Ibab  —  12ax  by  3a.  Ans.  5b  —  4x. 

8.  Divide  —  25a'^x  +  l^ax^  by  —  bax.       Ans.  ba  —  2>x. 

4.  Divide  lOab  +  Ibac  by  5a.  Ans.  2b  -f  3a 

5.  Divide  Wax  —  Mx  by  6x.  Ans.  ba  —  9. 

6.  Divide  Sx^  +  12x  by  4:x\  Ans.  2x  +  Bx-\ 

7.  Divide  Bbcd  +  126cx  —  Wc  by  36c?. 

-4ns.  d  +  4x  —  36 

8.  Divide  lax  +  lay  —  lad  by  —  la. 

Ans.  —  x  —  y  +  d 
0.  Divide  oax^  +  6x^  +  3ax  —  Ibx  by  3a:. 

A71S.  ax^  +  2x  +  a  —  5. 
10.  Divide  3a6*c  +  12ab'x  —  3a^6'  by  3a6«. 

Ans.  be  4-  46^a;  —  ab\ 

II.  Divide  25a26x  —  Iba^cx^  +  ba^bcx^  by  —  ba'^x. 

Ans.  —  56  +  3  ex  —  abcx. 

12.  Divide  20a^6^  +  15a'62  +  10a^6  +  5a  by  ba\ 

Ans.  4¥  4-  36^  -f  26  +  a-\ 

13.  Divide  21a  +  356  — 14  by  —  7.  ' 

Ans.  2  — 3a-— 56. 

Give  Oase  II.     Analysis.     Rule. 


58  ENTIRE  QUANTITIES. 

14.  Divide  —  12a'bc  +  Qacx""  —  Qab^c  by  —  oac. 

4ns,  4ab  —  ox'  +  25^ 

15.  Divide  6(a  +  x)  +  9(x  +  ij)  by  3. 

Ans.   2(a  +  x)  +  3(;?;  +  y). 

16.  Divide  12(a  +  ic)  —  3c(a  +  ^)  F  <^(a  +  x)  by  (a  +  i^). 

Ans.  12  —  3c  -f  d, 

17.  Divide  (a  +  c)'  —  (a  +  c)^  by  (a  -f  e). 

^ns.  (a  +  c)  —  (a  +  c)*. 

18.  Divide  12(a  — 6)  +  Qc(a  —  b)  +  2(a— 6)by  (a— 6). 

Ans.  12  +  6c  +  2 

19.  Divide  (m  +  n)x^  +  (m  +  ??)a'  +  (m  +  n.)c'  by  (m  -f  n) 

Ans.  x'  -f  a'^  4-  g\ 

20.  Divide  (a  1  by  +  2(a  +  &)  by  (a  +  ?>}. 

^ns.  a  +  Z>  +  2. 

Note. — Wlien  a  parenthesis  has  the  unit  1  for  both  coefficient  and  ex- 
ponent, and  is  connected  with  the  other  parts  of  the  algebraic  expres- 
sion by  -|-  or — » ^^  ^''^y  be  omitted;  thus,  (a-\-  b)  -\-  2,  is  the  same  as 
a  -\-  b  -\-  2.  But  when  a  parenthesis  having  the  minus  sign  before  it  is 
dropped,  the  signs  of  the  quantities  inclosed  must  all  be  changed  (63) ; 
thus,  a^  —  {a  —  x),  is  the  same  as  a^  —  a-\-  x 

21.  Divide  2a(a  +  c)  +  {a -^^  cy  by  (a  -f  c). 

Ans.  3a  +  c. 

22.  Divide  5c(3m  —  2c)  —  (Sm  —  2c)'  by  (S??!  _  2c). 

Ans.  1g  —  Sm. 

23.  Divide  (1  —  a:)  —  (1  —  x)'  by  (1  —  x).  Ans.  x 

CASE  ni. 

68.  To  divide  one  polynomial  by  another. 

Since  the  dividend  is  always  the  "product  of  the  divisor  hy 
the  quotient  souglit,  the  highest  power  of  any  letter  of  the 
dividend  must  be  the  product  of  the  highest  powers  of  the  same 
letter  in  tlie  divisor  and  quotient ;  and  the  inferior  powers  of 
this  letter  in  the  di  /idend,  must  be  the  products  of  inferior 
powers  in  divisor  and  quotient.     Hence   fhe   terms  of  both 


Give  Case  TIL 


DivisiaN.  59 

aimsor  and  dioidend  must  be  arranged  irv  the  order  of  the 

powers  of  one  of  the  letters. 

1.  Divide    2a'  -f  5aV  +  2a'b  —  Qab'  +  46*  by  a'  +  2ab 
\-  46'. 

OPERATION. 


a2  _j_  2ab  -(-  4b\  Divisor. 


2a'  —  2ab  -f  6^,  Quotient. 


Dividend,  2a*  -f-  2^36  4.  Sa^^s  _  g^^s  ^  4^>4 
2a*  4-  4a36  -f-  8a2i2 

1st  Rem.  —  2a^  —  Sa'^b'^  —  6ab^ 

—  2a4  —  ^a'^b'^  —  8a63 

2d  Rem.  a26*^  -f  2ab^  +  46* 

Analysis.  We  arrange  the  terms  of  both  divisor  and  dividend 
according  to  the  powers  of  a,  so  that  in  the  dividend  the  exponents 
of  this  letter,  taken  in  their  order,  are  4,  3,  2,  1 ;  and  in  the  divisor,  2, 
1.  Now,  according  to  the  principle  just  stated,  the  first  term  of 
the  dividend,  thus  arranged,  must  be  equal  to  the  first  terra  of  the 
divisor  multiplied  by  that  term  of  the  quotient  having  the  highest 
power  of  a;  we  therefore  divide  2a'*,  the  first  term  of  the  dividend, 
by  a^  the  first  term  of  the  divisor,  and  ol;)tain  2d^  for  the  first  of  the 
quotient.  We  next  miiiij^ly  the  whole  divisor  by  this  term  of  the 
quotient,  and  subtract  the  product  from  the  dividend,  bringing  down 
as  many  terms  as  are  necessary  for  a  new  dividend.  We  then  divide 
—  2a^6,  the  first  term  of  the  remainder,  by  d\  th-e  first  tetm  of  the 
divisor,  and  obtain  — 2ah  for  the  second  term  of  the  quotient.  We 
next  multiply  the  lohole  divisor  by  this  term  of  the  quotient,  and 
subtract  the  product  from  the  second  dividend,  and  obtain  a  second 
remainder  to  which  we  annex  another  term  of  the  dividend  for  another 
dividend.  Dividing  aW,  the  first  term  of  this  dividend,  by  a^  the 
first  term  of  the  divisor,  we  obtain  5^,  anothe.r  term  of  the  quotient. 
Lastly,  multiplying  the  whole  divisor  by  this  term  of  the  quotient, 
and  subtracting  the  product  from  the  last  dividend,  we  have  no 
remainder,  and  the  work  is  finished. 

From  this  example  we  derive  the  fQllowing 

E-HLE.  I.  Arrange  both  divisor  and  dividend  with  refe^ 
"ence  to  the  powers  of  one  of  the  letters. 

II.  Divide  the  first  term  of  the  dividend  by  the  first  term 
^f  the  divisor^  and  write  the  result  in  the  quotient. 


Give  analj'sis      Rule  for  the  division  of  polyuomiala. 


go  ENTIRE  QUANTITIES. 

III.  Multiply  the  whole  divisor^  by  the  quotitnt  thus  founds 
and  subt^ct  t/ie" product  from  the  dividend. 

I Y.  Arrange  the  remainder  for  a  new  dividend,  ivith  which 
proceed  as  before,  till  the  first  term  of  the  divisor  is  no  longer 
contained  in  the  first  term  of  the  remainder. 

Y.  Write  the  final  remainder,  if  there  be  any,  over  the 
divisor  in  the  form,  of  a  fraction,  and  the  entire  result  will 
e  the  quotient  sought. 


EXAMPLES   FOR   PRACTICBl 

2.  Divide  a^  +  2ax  +  x^hj  a  -\-  x.  Ans,  a  -f  a>, 

8.  Divide  a^  —  3a^?/  +  Say^—^y^  by  a  — y. 

Ans,  a?  —  2«2/  +  2/'« 

4.  Divide  a«  +  ha^b  +  hah^  +  b''  by  a  +  6. 

Ans.  a'  +  4a6  +  61 

5.  Divide  o:?  —  ^x^z  -f  z^\i^  x  —  z. 

Ans.  x^  —  ^xz  —  22^ . 

X  —  z 

6.  Divide  a^  +  la^b  +  "hab^  -f  H'  by  a^  +  ab  +  b\ 

Ans.  a  +  6 

7.  Divide  x""  —  ^x'  +  Tix  _  27  by  a?  —  3. 

Ans.  0?'  _  6^  +  9. 

8.  Divide  Ga;*  —  96  by  ^x  —  \% 

Ans.  x^  -f  2^2  _j.  4^  ^  g^ 

9.  Divide  ^a^  +  9a^  —  15a  by  Sa^  —  ^a. 

Ans.  2a^  +  2a  +  5. 
10.  Divide  25x^  —  x"  —  2x^  —  Sx  by  5^^  —  4x. 

Ans.  bx"  +  4x^  -\-  Sx  ^  2. 
11    Divide  ISa^  —  86^  by  6a  +  46.  .4ns.  Sa  —  26. 

12.  Divide  2^  —  19x'  -f  26^  _  16  by  ^  —  8. 

Ans.  2x^  _  3 jr  +  2 

13.  Divide  2/^ -f  1  by  2/ +  1.        Ans.  y^  —  V^+y^ — 2/  +  1. 

14.  Divide  y^  —  1  by  2/ — 1. 

Ans.  y^  +  y^  +  y^-h  y^  +  y  -hi. 

15.  Divide  x"^  —  a*  by  a?  —  a.  Ans,  x  +  a 


DIVISION.  61 

16.  Divide  6a'  —  3a'b  —  2a  +  6  by  3a'  —  1. 

Ans.  2a  —  6. 

17.  Divide  y^ — 3^/ V  +  3i/ V  —  x^  by  y^  —  dy^x  +  ^yx"^ — a^ 

Ans.  y'  +  oy^'x  -f  oyx'^  +  x^ 

18.  Divide  Ua'b^  —  2ba"b'  by  Sa'¥  +  bah\ 

Ans.  Sa'^b'  —  ^ab\ 

19.  Divide  2a*  —  2a;*  by  a  — -  ^. 

Ans.  2a'  +  2a^^+  2a^2  _^  2;2r\ 

20.  Divide  (a  —  ^)^  by  (a  —  xy,  A7is.  (a  —  xy 

21.  Divide  a'  —  od^x  +  ^ax^  —  a^  by  a  —  x. 

Ans.  c^  —  2ax  +  x^, 

22.  Divide  a'^  +  1  by  a  +  1. 

Ans.  a*  —  0^  -\-  Q^  —  a  -\-  \. 

23.  Divide  6«  —  1  by  6  —  1. 

Ans.  b"^  -\-b'  +  ¥  ^b^  +  b^-X. 
^4.  Divide  48a^  —  92a'a;  —  ^^ax"  +  lOOx^  by  8a  —  hx. 

Ans.  16a"^- —  4ax  —  20x1 

25.  Divide  4cZ*  —  9(Z^  +  6(Z  —  1  by  2cZ^'  +  3cZ  —  1. 

.4>26-.  2^'^  —  36Z  4-  1. 

26.  Divide  6a*  +  4a«x  —  9a'x^—  3ax^  +  2x*  by  2a'^  +  2t^ 
—  x^.  Ans.  ^a^  —  ax  —  2x1 

27.  Divide  3a*  —  Sa'^^'^  +  3aV  +  56*  ~  36V  by  a'^  —  b\ 

Ans.  3a'  —  56'  +  3c'. 

28.  Divide  2x'  +  "Ixy  +  6^/'  by  x  +  2y. 

Ans.  2x  +  3t/. 

29.  Divide  2mx  +  Swx  +  lOmn  +  ISn'  by  x  +  5n. 

Ans.  2m  +  3/1. 

30.  Divide  cZ*  —  3cZ'c  —  10c'  by  d'  —  5c.     Ans.  d'  +  2c. 

31.  Divide  m'  —  c'  +  2c2:  —  2'  by  m  +  c  —  z. 

Ans.  m  —  c  +  z. 

32.  Divide  y^  +  B2z^  hjy  +  2z. 

Ans.  2/*  —  2^2:  H-  42/'z'  —  82/2'  +  I62:*. 

33.  Divide  12(a  +  by  +  3(a  +  by  by  3(a  +  b). 

Ans.  4(a  -j-  6)'  +  a  +  6. 
84.  Divide  3c(m  —  5c)^  —  (m  —  5c)^  by  (in  —  5c)'. 

Ans.  8c  —  w. 
6 


Q2  ENTIRE  QUANTITIES. 

GENERAL  PRINCIPLES  OF  DIVISION 

09,  The  value  of  a  quotient  in  division  depends  apon  the 
relative  values  of  the  dividend  and  divisor;  and  the  sign  of  the 
quotient  depends  upon  the  relative  signs  of  the  dividend  and 
divisor.  Hence  any  change  in  the  value  or  the  sign  of  either 
dividend  or  divisor  must  produce  a  change  in  the  value  or  the 
sign  of  the  quotient ;  though  certain  changes  may  be  made  in 
both  dividend  and  divisor,  at  the  same  time,  that  will  not  affect 
the  quotient.  The  laws  that  govern  these  changes  are  called 
General  Principles  of  Division. 

CHANGE    OF    VALUE. 

70.  It  will  be  necessary  to  examine  only  those  changes  of 
value  produced  by  multiplying  and  dividing  the  dividend  and 
divisor. 

Let  us  take  abed  for  a  dividend,  and  ab  for  a  divisor ;  the 
quotient  will  be  cd,  and  the  operations  performed  upon  divi- 
dend and  divisor  will  affect  this  quotient  as  follows : 

Dividend.        Divisor.      Quotient. 

abed     -=r  ab    =     cd  ♦ 


1.  abcde  -^  ab    ==     cde 


(  Multiplying  the  dividend  by  e 
\      multiplies  the  quotient  by  e. 
r)      1^        _^     J     _  f  Dividing  the  dividend  by  d  di- 

(      vides  the  quotient  by  d, 
(  Multiplying  the  divisor  by  c  di- 
(      vides  the  quotient  by  c 

viding  the  divisor  by  b  mul 
ies  the  quotient  by  /;. 

5.  abcde  -^  abe  =  cd      I  Multiplying   both   terms   by  e 

I      does  not  alter  the  quotient. 

^      ,    J  ,  J       i  Dividing  both  terras  by  a  does 

^,     bed    -^     b     =  cd      \  :^  \ 

I      not  alter  the  quotient. 

\^hat  determines  the  value  of  a  quotient  in  division? 


8.  abed    -f-  abe  =     d 
4.  abed    -r-  a     =  bed 


(  Dividin 
\      tiplie 


DIVISION.  (53 

In  these  six  operations,  the  factors  employed  to  operate 
with  are  literal  quantities,  and  may  represent  any  numbers 
whatever ;  hence  the  results  are  general  truths  ;  they  may  be 
stated  as  follows  : 

Prin.  I.  Multiplying  the  dividend  multiplies  tJte  quotient 
and  dividing  the  dividend  divides  the  quotient.   (1  and  2.) 

Prin.  1 1.  Multiplying  the  divisor'  divides  tlie  quotient^ 
and  dividing  the  divisor  multiplies  the  quotient.   (3  and  4.) 

Prin.  III.  Multiplying  or  dividing  both  dividend  and 
divisor  by  the  same  quantity  does  not  alter  the  quotient 
(5  and  6.) 

Tl.  These  three  principles  may  be  embraced  in  one 

GENERAL   LAW. 

A  change  in  the  dividend  produces  a  like  change  in  the 
quotient;  but  a  change  in  the  divisor  produces  an  o1»posite 
change  in  the  quotient. 

CHANGE  OF  SIGN. 

72.  To  investigate  the  relative  changes  of  signs  in  division, 
let  it  be  remembered  that  when  the  divisor  and  dividend  have 
like  signs,  the  quotient  is  j:>Zms,  and  when  they  have  unlike 
signs,  the  quotient  is  minus.     Then 

l8^.  Suppose  the  divisor  and  dividend  have  like  signs ; 
if  either  of  the  signs  be  changed,  they  will  become  unlike,  an<I 
the  sign  of  the  quotient  will  be  changed  from  plus  to  mtruts. 

2d.  Suppose  the  divisor  and  dividend  have  unlike  signs ; 
if  either  of  the  signs  be  changed,  they  will  become  alike,  and 
the  sign  of  the  quotient  will  be  changed  from  minus  to  p/i^s. 

od.  Suppose  again  that  the  divisor  and  dividend  have  like 
signs  ;  if  both  signs  be  changed  at  once,  they  will  still  be  alike 
and  the  sign  of  the  quotient  will  remain  j^lus. 

Explain  the  principles  which  govern  changes  of  vahie  of  the  quotient 
in  division.  Repeat  Principle  I.  Prin.  IL  Prin.  III.  The  genera) 
law.  Explain  the  principles  which  govern  changes  of  signs,  of  divisor 
and  quotient. 


64  ENTIRE  QUANTITIES. 

4:th.  Suppose  again  that  the  divisor  and  dividend  have  un- 
like signs ;  if  both  signs  be  changed  at  once,  they  will  still 
be  unlike,  and  the  sign  of  the  quotient  will  remain  minus. 

These  results  may  be  embraced  in  two  principles,  as  follows  : 

pRiN.  I.  Changing  the  sign  of  either  dividend  or  divisor, 
changes  the  sign  of  the ^quotient. 

Prin.  II.  Changing  the  signs  of  both  dividend  and  divisor, 
does  noi  alter  the  sign  of  the  quotient. 

Note.  —  If  the  dividend  or  divisor  is  a  polynomial,  its  entire  value  is 
changed  by  changing  the  signs  of  all  its  terms. 

EXACT  DIVISION. 

73.  An  JExact  Division  is  one  in  which  the  quotient  has 
no  fractional  part. 

74.  From  the  rule  for  division  (66)  it  is  evident  that 
the  exact  division  of  one  monomial  by  another  will  be  im- 
possible :  — 

1st.  When  the  coefficient  of  the  divisor  is  not  exactly  con^ 
iained  in  the  coefficient  of  the  dividend. 

2d.  Wlien  a  literal  factor  has  a  greater  ex2:)onent  in  the 
divisor  than  in  the  dividend. 

od.  When  a  literal  factor  of  the  divisor  is  not  found  in 
the  dividend. 

75.  It  is  also  evident  (68)  that  the  division  of  one  poly- 
nomial by  another  will  be  impossible  : 

1st.  When  the  first  term  of  the  divisor  arranged  with  refer- 
ence to  any  one  of  its  letters  is  not  exactly  contained  in  the 
first  term  of  the  dividend  arranged  with  reference  to  the 
same  letter. 

Repeat  Prin.  I.  Prin.  II.  Define  an  exact  divisor.  Explain  under 
■what  circumstances  and  why  the  exact  division  of  one  monomial  by  an- 
other is  impossible.  Under  what  circumstances  and  why  is  on©  poly 
Hernial  not  au  exact  divisor  of  another  ? 


DIVISION.  g5 

2d  When  a  remainder  occurs ^  having  no  term  which  will 
exactly  contain  the  fii^st  term  of  the  divisor. 

76.  In  all  cases  where  exact  division  is  impossible,  the 
quotient  may  be  indicated  by  writing  the  dividend  above 
a  horizontal  line,  and  the  divisor  below,  according  to  defini- 
tion (7). 


RECIPROCALS,  ZERO  POWERS,  AND  NEGATIVE  EXPONENTS. 

77.  The  Reciprocal  of  a   quantity  is  1  divided  by  that 

quantity ;  thus  -  is  the  reciprocal  of  a ;  is  the  reciprocal 

^  a  X  —  y 

of  a;  —  ?/• 

78.  If,  in  the  division  of  powers,  we  conform  strictly  to  the 
rule  of  subtracting  the  exponent  of  the  divisor  from  the  expo- 
nent of  the  dividend,  then,  in  the  case  of  equal  powers,  the  ex- 
ponent of  the  quotient  will  be  0,  and  in  cases  where  the  divisor 
is  the  higher  power,  the  exponent  of  the  quotient  will  become 
negative. 

70.  To  explain  the  import  of  a  cipher  when  used  as  an 
exponent,  we  observe  that  the  quotient  of  any  quantity  divided 
by  itself  is  1 ;  consequently,  when  the  divisor  and  dividend  are 
like  powers  of  the  same  quantity,  we  may  have  two  expressions 
for  the  quotient ;  thus 

-  =  a^  ^  =  a^.  or  -=  1; 
a  a         ' 

^  =a— "=a^or\=l. 
a"*  a"* 


Therefore,  (Ax.  7),       /,      a^  =  1. 


I 


In  the  cases  previously  stated  how  may  the  quotient  be  written  ? 
Define  the  reciprocal  of  a  quantity.  In  Division  when  will  the  exponent 
of  the  quotient  be  0?  When  negative?  Explain  why.  What  is  the 
value  of  any  quantity  whose  exponent  is  0  ?     Why  ? 

6  *  E 


OQ                                     ENTIRE  QUANTITIES.  \ 

But  a  may  represent  any  quantity  whatever.     Hence,  | 

Any  quantity  having  a  cipher  for  an  exponent  is  equal  to  j 

unity,  \ 

Note. — When  a  quantity  "with  a  cipher  for  an  exponent  is  a  factor  in  j 

t-n  algebraic    expression,  it  may  he   suppressed  without   affecting  the  ; 

value  of  the  expression ;  yet  it  is  frequently  retained  in  order  to  indicate  I 

the  process  by  which  the  result  was  obtained.  ; 

80.  To  show  the  signification  of  negytive  exponents,  let  us  i 

divide  a^  by  a^  by  taking  the  difference  of  the  exponents  ;  thus,  j 

a^  -^  a^  =  a^'"^  =  a~^,  j 

But  the  value  of  the  quotient  will  not  be  altered  if  we  divide  j 

both  dividend  and  divisor  by  a^  (70,  III)  ;   thus,  i 

1  A 

a^  -f-  a'  =  1  -f-  a^  =  -.  j 

I 

\ 

These  quotients  being  equal,  we  have  l 


a' 


] 


This  principle  may  also  be  illustrated  as  follows.     Since  the 
zero  power  of  any  factor  is  1,  we  may  have 

„  =  a     '  =  a-^  or      =  -;  hence,  a-""  =  -  • 
a^  a^       a^  a^ ' 

a'         n  a'      1      ,  1  / 

—  =  a^-""  =  fl-"", or-  =  —  :  hence, a~'"=  --.  / 


From  these  illustrations  we  deduce  the  following  inference : 

Any  quantity  having  a  negative  exponent  is  equal  to  the 
reciprocal  of  that  quantity  with  an  equal  positive  exponent. 

What  do  negative  exjronents  signify  f     What  relation  do  they  bear  to 
reciprocals. 


DIVISION.  67 


FACTORING. 


I 


81.  The  Factors  of  a  quantity  are  those  quautities  which, 
being  multiplied  together,  will  pi;pduce  the  given  quantity."" 

8S.  A  Composite  Quantity  is  one  that  may  be  produced 
by  the  multiplication  of  two  or  more  factors.  A  composite 
quantity  is  exactly  divisible  by  any  of  its  factors. 

8JI.  A  Prime  Quantity  is  one  that  cannot'  be  produced  by 
the  multiplication  of  two  or  more  factors,  and  is  divisible  only 
by  itself  and  unity. 

Several  quantities  are  prime  to  each  other  when  they  have 
^no  common  factor,  or  when  no  quantity  except  unity  will  divide 
them  all. 

CASE   I. 

84,  To  factor  a  monomial. 

The  prime  factors  of  a  purely  algebraic  quantity,  consisting 
of  a  single  term,  are  visible  to  the  eije  ;  and  this  is  one  of  the 
principal  advantages- of  an  algebraic  expression.  Algebraic 
quantities  are  factored  by  inspection  or  by  trial,  the  same  as 
numbers  in  arithmetic. 

1.  What  are  the  prime  factors  of  6a'6V? 

OPERATION. 

r-   __  9  V  3  Analysis.     The  prime  f^iotors    of  6 

,  are  2  and  3 ;  the  exponents  show  that 

^  a  18  taken  6   times  as  a  factor  in   the 

\           0=0X0  given   term,  b  twice,  and  c  once ;   and 

c  —  c  ^ Qa^b'^c  =  2  X  oaaabbc.    Hence, 

2    X  oaaabbc 

Utile.  Resolve  the  numeral  coefficient  into  its  prime  fac- 
I  tors,  and  write  each  letter  as  many  times  as  there  are  units  in 
l^.its  exponent. 

Define  factors.  A  composite  quantity.  A  prime  quantity.  Quantities 
prime  to  each  other      Wb-^t  is  Case  I  ?    Give  analysis.     Rule. 


68  ENTIRE  QUANTITIES.  f 

EXAMPLES   FOR  PRACTICE. 

2.  Resolve  lOx^y^  into  its  prime  factors. 

Ans.  2  X  bxxyyy, 

3.  Resolve  15m^c*  into  its  prime  factors. 

4.  Resolve  24j;V  into  its  prime  factors. 

Ans.  2  X  2  X  .2  X  ^ppppziL 

5.  Resolve  75a^b^c(P  into  its  prime  factors. 

6.  Resolve  2Qm*x^yz  into  its  prime  factors. 

Ans,  2  X  l^mmmmxxyz. 

CASE  II. 

85.  To  resolve  a  polynomial  into  two  factors,  a 
monomial  and  a  polynomial. 

Polynomials  may  be  factored  by  inspection  under  certain 
conditions.  If  the  terms  have  a  common  factor,  the  quantity 
may  be  separated  into  two  factors,  a  monomial  and  a  polynomi-aL 

1.  Resolve  2ax  —  2am  +  ^az  into  its  factors. 

Analysis.    Since  2ft  is  a  factor  common  to  all  the  terms,  we  divide 
by  this  factor,  and  obtain  for  a  quotient,  x  —  m  -{■  Sz^  which  is  the 
other  factor  of  the  given  quantity ;  or,  2ax  —  2am  +  Qaz  =  2a{x 
m  +  32;).     Henoe, 

Rule.  Divide  by  the  greatest  factor  common  to  all  the 
terms,  inclose  the  quotient  in  a  parenthesis,  and  write  the 
divisor  as  the  coefficient. 

EXAMPLES   FOR  PRACTICE. 

2.  Find  the  factors  of  ax  +  bx.  Ans.  x(a  +  ft). 

3.  Find  the  factors  oi  x  +  ax,  Ans.  x(l  +  a). 

4.  Resolve  am  +  an  +  ax  into  its  factors. 

Ans.  a(m  +  n  +  x). 
6.  Resolve  bc^  —  bcx  —  bey  into  its  factors. 

Ans.  bc(c  —  X  —  y). 

What  h  Catie  II  ?     Give  analysis.     Rule. 


DIVISION.  QQ 

6.  Resolve  4^7'  —  6xy  into  its  factors. 

Ans,  2x{2x  —  Sy). 
1,  Factor  a'^Z)' —  a'c  —  2a'd  Ans.  a\¥ —■  c -^^  2d). 
S.  Factor  3m^z  —  4:my  +  2c^m. 

Ans.  m(Smz  —  42/  +  2c'^). 
9.  Factor  12c^bx^  —  15cV  —  6cV?/. 

^ns.  3cV(4c26  —  bcx  —  2y). 

10.  Factor  ex  —  3cxz  +  aa?^  Ans.  cx(l  —  Sz  +  x). 

Note.  —  It  may  happen  that  a  portion  of  a  polynomial  can  be  factored 
when  there  is  no  factor  common  to  all  the  terms. 

11.  Factor  x^  +  2bx  —  66a  Ans.  x"  +  2h{x  —  3c). 

12.  Factor  a^n  +  ma  +  m6.  Ans.  a^n  +  m(a  +  b) 

13.  Factor  aa;^  +  3a^a7  +  6^' +  36^^. 

(ax(x  +  3a)  +  bx{x  -f  36)  ;  or, 
^^^-    lx\a  +  b)+  Zx(a^  +  b'). 

14.  Factor  a'  +  a'b  +  a6'  +  b\ 

Ans    W-\-ab-\-¥)  +  b'',  or, 
*   la^  +  b(o?  +  ab  +  6^). 
16.  Factor  a^z"  +  ^^^2'  +  x'^z  +  a;2l 

i^^2;2(^  +  z)  +  xzQjc  +  z)  ;  or, 
(x'^z'^  +  572;)  (^  +  z)  ;  or, 
xz(xz  +  1)  (57  +  z). 
16.  Factor  ax  -\-  ay  +  bx  +  by. 

L  .      f^c^  +  y)  +  K^  +  y)]  or, 

^''^-  |(a  +  6)  (a;  +  2/). 

CASE  in. 

86.  To  resolve  a  trinomial  into  two  equal  binomial 
factors. 

A  trinomial  may  be  resolved  into  two  binomial  factors  when 
tw;  of  its  terms  are  perfect  squares  and  positive,  and  the  other 

What  is  Case  III  ? 


YO  ENTIRE  QUANT  [TIES 

term  is  twice  the  product  of  their  square  roots,  and  either  po- 
sitive or  negative. 

1.  Factor  a'  +  2ac  -f  g\ 

Analysis,  d^  is  the  square  of  a,  d*  is  the  square  of  c,  and  2ac  ia 
twice  the  product  of  a  and  c;  and  since  a^  -f  2ac  -j-  c^  is  the  sum  of 
the  squares  of  a  and  c  plus  twice  their  product,  it  must  be  the  square 
of  a  -f-  c  (60) ;  or  a'  -f  2<xc  4-  c"  =  [a  -f  c)  {a  +  c).     Hence, 

KuLE.  Connect  the  square  roots  of  the  two  squares  by  the 
sign  of  the  other  term,  and  write  the  result  twice  as  a  factor. 

EXAMPLES   FOR  TRACTICE. 

2.  Resolve  a^  -\-  2ax  +  x^  into  its  factors. 

Ans,  {a  +  x)  (a  +  x). 

3.  E-esoIve  a^  —  2ax  -f-  x^  into  its  factors. 

Ans,  {a  —  x)  (a  —  x). 

4.  Resolve  A^  —  2AB  -\.  B^  into  its  factors. 

Ans,  (A^B)  (A  —  B). 

5.  Resolve  P'  +  2PQ  -j.  Q^  into  its  factors. 

Ans,  (P  -{-  Q)  (P  +  g). 

6.  Resolve  9a^  4-  12a6  +  46^  into  its  factors. 

^ns.   (3a  +  26)  (3a  +  2b). 

7.  Resolve  4m^  —  4m  +  1  into  its  factors. 

Ans.  (2m— 1)  (2m  — 1). 
8    Resolve  4c^  —  4c(Z  +  d^  into  its  factors. 

^72s.  (2c  — d)  (2c  — ci). 
9.  Factor  9m'  +  12m  +  4.     ^Ins.  (3??i  -f-  2)  (3m  +  2). 

10.  Factor  1  —  12;?  +  36z\         Ans.  (1  —  6z)  (1  —  6^). 

11.  Factor  aV  —  2ac  +  1.         Ans.  (ac  —  1)  (ao  —  1). 

12.  Factor  a*  +  2ax^-\-  a?x^.     Ans.  (x^  -j,  a:»)  (x^  ^  ax). 

13.  Factor  </«  -—  2\fz^  +  2«  ^ns.  (^  —  z^)  (1/  —  s^. 

Give  analysis      Rule. 


DIVISION.  71 

CASE  IV. 

87,  To  resolve  a  binomial  into  two  binomial  factors. 

A  binomial  may  be  resolved  into  two  binomial  factors,  when 
both  of  its  terms  are  perfect  squares,  and  have  contrary  signs. 

1.  Factor  a^— 61 

Analysis,  d^  is  the  square  of  a,  W  is  the  square  of  6 ;  and  since 
a}  —  V  is  tlie  difference  of  the  squares  of  a  and  6,  it  must  be  equal  to 
the  product  of  the  sum  and  difference  of  a  and  5  (61) ;  or  a^  —  lp-^=^ 
^a -f-  h)  (a  — 6).     Hence, 

Rule.  Write  the  sum  arid  difference  of  the  square  roots 
of  the  two  given  terms,  as  two  binomial  factors. 

EXAMPLES    FOR   PRACTICE. 

2.  Factor  x^  —  y\  Ans,  {x  +  ?/)  {x  —  y) 

3.  Factor  m^  —  nl  Ans.  (m  +  n)  (in  —  n). 

4.  Factor  y"^  —  Az^  Ans.  (jy  +  22)  (y  —  2z). 

5.  Factor  Aa'  —  db\  Ans.  (2a  +  36)  (2a  —  36). 
G.  Factor  25c'^  —  1.  Ans.  (5c  -f  1)  (5c  —  1). 

7.  Factor  o()c*d'  —  lG??t^ 

Ans.  (Qc'd  +  4m^)  (Gc^cZ  — 4m^). 

8.  Factor  9a'cV  -—  1.      Ans.  (Sac^x  +  1)  (Sac'x  —  1). 

9.  Factor  a.'^z^  —  a^y'\  Ans,  (az  -f-  ay)  {az  —  ay). 

10.  Factor  a'  —  &.  Ans.  (a'  -f-  c'^  (a  +  <?)  (a  — -  c). 

11.  Factors*  —  y\  Ans.  {x^  +2/0  (•^  +  2/)  (^  —  2/)' 

12.  Factor  x^ — z^. 

A71S.  (x'  +  z')  (x'  +  z')  (x  +  2)  (a;  —  z). 
13    Factor  m^«  —  c^^ 

Ans.  (m®-f-c^)  (m^-}-c*')  (ni'^i-c^)  (niic)  (m — c) 
14.   Factor  c^'^  —  1. 

.4ns.  (c^«  +  l)  (c«+l)  (c'-j-l)(€'  +  l)  (c  +  l)  (c— 1) 
15,.  Factor  c'^c  —  c.  Ans.  c(a  +  1)  (a  —  1). 

16.  Factor  a^c'  —  c\  Ans.  c\a  4-  1)  (a  —•  1). 

17..  Factor  xYz"  —  xY-  Ans.  xHf  (z  4.  1)  (z  —  1). 

What  is  Case  IV?     Gire  analysis.     Rule. 


X2  ENTIRE  QUANTITIES. 

18.  Factor  a;' —  a;. 

Ans.  x(x'  -\-  1)  (x^  +1)  (x  +  1)  (x—  1). 

19.  Factor  m^  —  m^ 

A  (\    ,  Ans.  m^m^  +  1)  (m  +  1)  (m  — 1). 

GREATEST  COMMON  DIVISOR. 

88.  A  Common  Divisor  of  two  or  more  quantities  is  a 
quantity  that  will  exactly  divide  each  of  them. 

89.  The  Greatest  Common  Divisor  of  two  or  more  quanti- 
ties is  the  greatest  quantity  that  will  exactly  divide  each  of 
them. 

90.  It  is  evident  that  if  two  or  more  quantities  be  divided 
by  their  greatest  common  divisor,  the  quotients  will  be  prime 
to  each  other. 

1.  What  is  the  greatest  common  divisor  of  ^a^Wcd^  ^Sa^^c^x, 
and  Ua'b'cd'  ? 

OPERATION. 

ia'b^cd  =  2^  X  a"  X  h^  X  c    X  d 

^Sa'b'c'x  =  3    X  2*  X  a*  X  6^  X  c^  X  x 

Ua'b'cd'  =  3x2^  x  a'  X  b'  X  c   x  d^ 

~Ia^     ^  2^  X  a^  X  b'  X  G 

Analysis.  We  resolve  the  quantities  into  their  component  factors^ 
and  write  all  the  powers  of  the  same  factor  under  each  other.  By 
inspection  we  perceive  that  all  the  quantities  contain  at  least  the 
second  power  of  2,  and  we  write  2^  underneath  as  a  factor  of  the 
greatest  common  divisor  sought ;  all  the  quantities  contain  at  least 
the  second  power  of  a^  and  we  write  a^  underneath;  all  the  quanti- 
ties contain  at  least  the  second  power  of  b  and  the  first  power  of  c, 
and  we  write  these  factors  underneath ;  and,  since  these  are  all  the 
common  factors,  their  product,  4a^6^c,  must  be  the  greatest  common 
titiaor  of  the  given  quantities. 

2.  What  is  the  greatest  common  divisor  of  3ac^  (x*  —  c*), 
and  a^cx'^  —  aV  ? 

Define  a  common  diyisor.  The  greatest  common  divisor.  Give  analy- 
elB  of  Example  1 


DIVISION.  78 

OPEBATION. 

Bad"  (.'»*  — c*)  =  3  X  a  X  c^  X  (a?^  +  c")  X  (a?  +  c)  X  (ar  — c) 
d^'x^  —  aV  =         a?x  c  X  (x  +  d)  X  (x  —  c) 

cc(^  —  c^)  =         axcx  (a7  4-c)x  (a?  —  c) 

Analysis.  Resolving  the  quantities  into  factors  as  before,  we 
rftadily  perceive  that  the  only  common  factors  are  a,  c,  (a;  -f-  c),  and 
(t  —  c) ;  and  the  product  of  these  factors,  ac  (x^  —  c^),  must  there- 
fore be  the  greatest  common  divisor  sought. 

From  these  examples  we  deduce  the  following 

Rule.  I.  Resolve  the  given  quantities  into  factors^  and 
write  all  the  powers  of  the  same  factor  under  each  other, 

II.  Ilultiply  together  the  lowest  power  of  each  common 
factory  and  the  product  will  be  the  greatest  common  divisor 
%ought. 

EXAMPLES  FOR  PRACTICE. 

3.  What  is  the   greatest  common  divisor  of  4a V,  and 
I0a5c^?  Ans.  2ac\ 

4.  What  is  the  greatest  common   divisor  of  Zahx^,  and 
12abx^zl  Ans.  ^abx", 

5.  What  is  the  greatest  common  divisor  of  ^a^H^ar^z^^  and 
8a^^^2^?  Alls,  ^a^x^z^. 

6.  What  is  the   greatest  common  divisor   of   4:am'^y^2^, 
12m^2®,  and  IQa^mh^  ?  Ans.  ^ni^z^. 

7.  What  is  the  greatest  common  divisor  of  6aVd^  12aVc^'^, 
^a^c^d\  and  24:a'c'dm  ?  '  Ans.  Ba^c'd. 

8.  Find  the  greatest  common  divisor  of  a^  —  6^  and  a^  — 
2ab  +  ¥.  Ans.  a  —  b. 

9.  Find  the    greatest   common  divisor  of   a"^  —  c\   and 
a*  +  2ac  +  c\  Ans.  a  +  c. 

10  Find  the  greatest  common  divisor  of  m^  —  2m,  and 
2mn^  —  4:n\  An&.  m  —  2. 

Of  E^^ampild  2.    Eule. 


74  ENTIRE  quantities;. 

11.  Find  the  greatest  common  divisor  of  ax^  —  ay^^  aw?x  — 
am^ijj  and  aV  —  2a^xy  +  a^.  Ans.  a{x  —  y). 

12.  Find  the  greatest  common  divisor  of  16a^  —  c^  and 
16a*  —  8ac  +  c^  Ans.  Aa  —  c. 

13.  Find  the  greatest  common  divisor  of  Sa^b  —  9a^c  — 
ISa^mz,  and  b^c  —  Sbc"^  —  Qbcmz,  Ans,  b  —  3c  —  Qmz, 

LEAST  COMMON  MULTIPLE. 

91.  A  Multiple  of  a  quantity  is  another  quantity  exactly 
divisible  by  it. 

Note.  —  If  a  quantity  be  multiplied  by  any  factor,  the  result  is  pro- 
perly called  a  multiple  of  that  quantity,  and  this  is  the  real  signification 
of  the  word  multiple.  But  the  product  is  always  divisible  by  the  multi- 
plicand ;  hence  the  definition  as  given  above. 

92.  A  Cominon  Multiple  of  two  or  more  quantities  is  one 
which  is  exactly  divisible  by  each  of  them. 

93.  The  Least  Common  Multiple  of  two  or  more  quantities 
is  the  least  quantity  exactly  divisible  by  each  of  them. 

94.  It  is  evident  that  one  quantity,  to  be  divisible  by  seve- 
ral other  quantities,  must  contain  all  the  factors  in  each  of  the 
given  quantities. 

1.  What  is  the  least  common  multiple  of  Sa^x^y,  and 
12a»5'x? 

OPERATION. 

SaVy     =  2^  X  a^  X  x^  X  y 

l2aWx     =  3  X  2^  X  q"  X  x  x  6» 

24a»6V2/  =  3  x  2»  x  a'h<  x"  x  y  X  b* 

Analysis.  We  resolve  the  given  quantities  into  their  component 
factors,  and  write  the  powers  of  each  separate  letter  or  factor  under 
each  other.  The  different  prime  factors  are  3,  2,  a,  or,  y,  and  b ;  and 
the  least  common  multiple  must  contain  not  on'y  each  of  these,  but 
the  highest  power  of  each  that  is  contained  in  the  given  quantities, 

Define  a  Multiple.  A  common  multiple.  The  least  common  multiple. 
When  one  quantity  is  a  common  multiple  of  Several  others,  "what  must 
U  contain?     Gm  analy'sis  of  Example  1. 


DIVISION 


75 


otherwise  it  will  not  contain  all  the  factors  of  the  given  quantities. 
By  inspecting  the  powers  of  each,  the  highest  are  the  1st  power  of  3^ 
3d  power  of  2,  3d  power  of  a,  2d  power  of  a;,  1st  power  of  ?/,  and  3d 
power  of  b ;  and  their  product  is  24a^6'^a?^y,  the  least  common  mul- 
tiple required.     Ilence,  the 

Rule  I.  Resolve  the  given  quantities  into  factors,  and 
write  all  the  powers  of  the  same  factor  under  each  other. 

II.  Multiply  tocjether  the  high  est  power  of  every  factor,  and 
the  product  will  he  the  least  common  multiple  sought 

EXAMPLES   FOR   PRACTICE. 

2.  Find  the  least  common  multiple  of  oaWc,  bab%  abd^^ 
and  I5a'¥c.  Ans.  Ua'b'cd'. 

3.  Find  the  least  common  multiple  of  Qxy,  9x*z,  ^x^y% 
and  .r*z.  Ans.  ISjj^'-^z. 

4.  Find  the  least  common  multiple  of  2mn,  3mV,  Qmz^, 
and  ^mnz,  Ans.  12m'^nz\ 

5.  Required  the  least  common  multiple  of  27a,  156,  dab, 
and  3a-.  Ans.  135a'6. 

6.  Find  the  least  common  multiple  of  (a^  —  x^),  4(a  —  x), 
and  (a  +  x),  Ans.  4(a'  —  x^ 

7.  Required  the  least  common  multiple  of  a\a  —  x),  and 
ax\a^  —  x').  AnS'  ^^'V  (a^  —  x^). 

8.  Required  the  least  common  multiple  of  x'(x  —  y),  a*x\ 
£indl2axy\  Ans.  V2a*xY\^  —  y)- 

9.  Required  the  least  common  multiple  of  10aV(a  —  6) 
16af>(a  +  b),  and  12(a'^  —  6").  A^is.  60aV(a^  —  b'). 

10.  What  is  the  least  common  multiple  of  m*  —  1,  m*  — 
2?/i  -h  1,  and  m'  +  2m  +  1  ?  An,<i.  7?i«  —  m'  —  m'  +  1. 

11.  What  is  the  least  common  multiple  of  ^^ — y^,  x'^y  —  xy*^ 
smdjc^y  +  xif  t  Ans.  a^y  —  xif. 

12.  What  is  the  least  common  multiple  of  m^  —  4,  zm  —  2z, 
and  m^  +  27?i  ?  Ans,  zm^  —  4zm. 

Give  Rule.  , 


^Q  FRACTIONS. 


FRACTIONS. 

95.  The  word  Fraction  relates  to  a  certain  mode  or  form 
of  indicating  division ;  and  fractional  forms  have  precisely  tlio 
same  signification  in  Algebra  as  in  Arithmetic. 

96.  A  Fraction  is  a  quotient  expressed  by  writing  the 
dividend  above  a  horizontal  line,  and  the  divisor  below ;  thus, 

—  is  a  fraction,  and  is  read,  a  divided  by  b. 
b 

97.  The  Denominator  of  a  fraction  is  the  quantity  below 
the  line,  or  the  divisor. 

98.  The  [Kfumerator  of  a  fraction  is  the  quantity  above  the 
line,  or  tlie  dividend. 

99.  Since  a  quantity  is  divided  by  dividing  any  one  of  its 

factors,  we  have  --  = — 7 —  =  --  x  a ;  hence,  a  fraction  is  equal 

to  the  reciprocal  of  its  denominator  multiplied  by  its  nume- 
rator. 

100.  An  Entire  Quantity  is  an  algebraic  expression  which 
has  no  fractional  pait ;  as,  3a,  or  a;  —  Sy^. 

101.  A  Mixed  Quantity  is  one  which  has  both  entire  and 
c  Sg 


fractional  parts  ;  as,  a  H — ,  m 


a'  l  —  b' 

SIGNS. 

10^.  Each  term  in  the  numerator  and  denominator  of  a 
Taction  has  its  own  particular  sign,  distinct  from  the  real  sign 

A  the  fraction.   Thus,  in  the  fraction,  - — -^^-^ — ^,  the  signs 

X  y  —  xy 

Define  a  fraction.  A  denominator.  A  numerator.  A  fraction  is 
equal  to, the  reciprocal  of  what?  'Define  an  entire  quantity.  A  mixed 
^aairtity. 


GENERAL  PRINCIPLES.  ^f 

of  a  part  of  the  terms  only  are  expressed.  If  no  sign  is  pre- 
fixed to  the  first  term  of  a  numerator  or  denominator,  the  plua 
sign  is  understood. 

Il>3.  The  Apparent  Sign  of  a  fraction  is  the  sign  written 
before  the  dividing  line,  to  indicate  whether  the  fraction  is  to 

be  added  or  subtracted ;  thus,  mm  -\ ,  the  apparent 

sign  of  the  fraction  is  plus,  and  indicates  that  the  fraction  is  to 
be  added  to  m. 

104.  The  Real  Sign  of  a  fraction  is  the  sign  of  its  numeri- 
cal value,  when  reduced  to  a  monomial,  and  shows  whether  the 
fraction  is  essentially  a  positive  or  a  negative  quantity ;  thus, 

in  the  last  fracti6n,       "" — '-,  let  a?  =  2,  and  a  =  12  ;    then 
a  —  X 

a^  —  ax      4-^12x2      —20  _     _  ^.  .. 

=  — r- —  =  — ,rj-  =  —  2.    Hence,  the  real  sign 

a  —  X  12  —  2  10 

of  this  fraction  is  minus,  though  its  apparent  sign  ispZi^s 


GENERAL  PRINCIPLES  OF  FRACTIONS. 

10«i«  Since  fractions  indicate  division,  all  changes  in  the 
numerator  and  denominator  of  a  fraction  will  afTect  the  value 
and  sign  of  that  fraction  according  to  the  laws  of  division ; 
and  we  have  only  to  modify  the  language  of  the  General  Prin- 
ciples of  Division  (70),  by  substituting  the  words  numerator^ 
denominator^  and  fraction^  for  the  words  dividend^  diviso?^ 
and  quotient,  and  we  shall  express  the  laws  governing  the 
changes  in  the  value  and  sign  of  a  fraction. 

CHANGE   OF   VALUE. 

PRIN.  I.  Multiplying  the  numerator  multiplies  the  frao- 
Hon,  and  dividing  the  numerator  divides  the  fraction. 

Define 'the  apparent  sign  of  a  fraction.  The  real  sign.  Fractions 
alvrays  indicate  what  ?  Adapt  the  general  principles  of  Division  to  frao- 
tions.     Repeat  the  priiwiples  that  govern  change  of  value. 

7* 


78  FRACTIONS. 

Prin.  II.  Multiplying  the  denominator  divides  the  frac- 
tion, and  dividing  the  denominator  multiplies  the  fraction. 

PuiN.  III.  Multiplying  or  dividing  both  numerator  and 
denominator  by  the  same  quantity  does  not  alter  the  valu:?  of 
the  fraction, 

106.  These  three  principles  may  be  embodied  in  one 

GENERAL   LAW. 

A  change  in  the  numerator  pi^oduces  a  like  change  in  the 
value  of  the  fraction  ;  but  a  change  in  the  denominator  pro- 
duces  an  orposiTE  change  in  the  value  of  the  fraction. 

CHANGE   OF   SIGN. 

107.  Prin.  I.  Changing  the  sign  of  either  numerator  or 
denominator,  changes  the  real  sign  of  the  fraction. 

Prin.  II.  Changing  the  signs  of  both  numerator  and  da- 
nominator  at  the  same  time^  does  not  alter  the  real  sign  of  the 
fraction. 

Prin.  III.  Changing  the  ap)parent  sign  of  the  fraction 
changes  the  real  sign. 

reduction. 

108.  The  Reduction  of  a  quantity  is  the  operation  of 

changing  its  form  without  altering  its  value. 

CASE   I. 

100.  To  reduce  a  fraction  to  its  lowest  terms. 

A  fraction  is  in  its  lowest  terms  when  its  numerator  and 
denominator  are  prime  to  each  other. 

Uah'c 
1.  Reduce  ^^  2/^2  to  its  lowest  terms. 

Apply  the  general  law.  The  principles  that  govern  change  of  signs, 
Define  Reduction-     What  is  Case  I? 


REDUCTION.  «fg 

OPERATION.  Analysis.      If  we   divide  both 

numerator  and  denominator  of  this 

i4aoc     *-0^  fraction  by  the  same  number,  its 

2ld^bc^  oao  value  will   not   be   changed  (105, 

III)  ;  and  if  wo  divide  by  the  great- 
est common  divisor,  the  quotients  will  be  prime  to  each  other  (90), 
and  consequently  the  fraction  will  be  in  its  lowest  terms.  By  inspec- 
tion we  find  labc  to  be   the  greatest   common   divisor;    and  divi- 

ding  both  terms  by  this  quantity,  we  have  o— ,  the  answer. 

d^x  +  ax"^ 

2    Reduce  — r. ;;-  to  its  lowest  terms. 

a'  —  x^ 


OPERATION. 

a^x  -f  ax^  ax  (a  -^  x)  ax 


a^  —  x^  (a  —  x)  (a  +  x)         a  —  x 

Analysis.  We  first  resolve  the  numerator  and  denominator  into 
their  prime  factors,  and  then  cancel  the  common  factor  (a  -f  x),  and 

we  have ,  the  answer. 

a  —  X 

From  these  examples  we  deduce  the  following 

Rule.  Divide  both  rrumerator  and  denominator  by  their 
greatest  common  divisor.     Or, 

Eesolve  the  numerator  and  denominator  into  their  prim^ 
/actors,  and  cancel  all  those  that  are  common, 

EXAMPLES   FOR   PRACTICE. 

8.  Reduce  —-7  t^  its  lowest  terms.  Ans,  ot. 

ioao  00 

,    ^   ,       14aVv       ..   ,        .  .  ^       2at/ 

4.  Reduce  r-. — -^  to  its  lowest  terms.  Ans,  -o^. 

zlax^  o 

Give  analyses.     Rule. 


go  FRACTIONS. 

5.  Reduce    ^^  ^     ,  to  its  lowest  terms.       Ans,  -zr^ — . 

bSa^xy^  liy 

6.  Reduce  ^r^rrz — pr-r-  to  ^^s  lowest  terms. 

17a^  — 21fl5 

7.  Keduce  jtt — :— ^  to  its  lowest  terms. 

8.  Reduce  — ^ 77-  to  its  lowest  terms.    Arts,     .,  .  ,^ 

a;*  —  6*  07^  +  6*^ 

^2 1  ^ ][ 

9.  Reduce • —  to  its  lowest  terms.         Ans,  . 

xy  +  y  y 

ex   I   cot^ 

10.  Reduce ■ — 7—  to  its  lowest  terms. 

acx  +  ahx  c  4-  cr 

Arts.   -^-^. 
ac  +  CO 


11.  Divide  a^y^  +  a?V  by  aoc^y  +  a^y*.  ^^«.  — . 


12.  Divide  4a  +  46  by  2a'  _  261  Ans. 


xy 
I 

2 


a  —  b' 


8 


13.  Divide  n^  —  2n^  by  n^  —  4n  +  4.  ^ns.   -^, 

14.  Reduce  — 5 =-  to  its  lowest  terms, 

15.  Reduce    ^  .  ^ — ; — 5  to  its  lowest  terms. 

x^  +  2ca7  +  G^ 

16.  Reduce  ^^-^ ~  to  its  lowest  terms. 

or  —  a' 

17.  Reduce     ,  ,     ,^  to  its  lowest  terms. 

ar  4-  ary^ 


REDUCTION.  81 

CASE  n. 

11©.  To  reduce  a  fraction  to  an  entire  or  mixed 
quantity. 

(lb  -\-  X 

1.  Reduce  — ^ — -  to  a  mixed  quantity. 

OPERATION.  Analysis.     Since  the  value  of 

-    .  the  fraction  is  the  quotient  of 

"^      =  a  -| t^®  numerator  divided  by  the  de- 

^  ^  nominator,  we  perform  the  divi' 

sion  indicated,  and  obtain  a  for 

X 

the  entire  part  of  the  quotient,  and  H-  t  for  the  fractional   part. 
Hence,  the 

KuLE.  I.  Divide  the  numerator  by  the  denominator  as 
far  as  possible,  for  the  entire  part, 

II.    Write  the  remainder  over  the  denominator,  and  annex 

the  fraction  thus  found  to  the  entire  2^cirt,  with  its  proj^er 

sign. 

Note.  —  If  any  term  be  found  in  the  numerator,  whose  literal  part 
is  exactly  divisible  by  some  term  in  the  denominator,  and  having  a 
greater  coefficient  than  this  term  of  the  denominator,  the  reduction  will 
be  possible ;  otherwise,  it  will  be  impossible. 

EXAMPLES   FOR  PRACTICE. 

19         a^  4-  bx 

2.  Reduce  -^  and to  mixed  quantities. 

bx 

Ans.  2  J  and  a  -i . 

^  a 

8.  Keduce  — to  a  mixed  quantity. 

Ans.  5a  -\ • 


2a* 26' 

4.  Reduce r— ■  to  an  entire  quantity. 


Ans.  2a  +  26, 


Give  Case  II.     Analysis.     Rule. 
P 


g2  FRACTIONS. 

1 5(/S 2.x 

5.  Reduce  — — r-^-- ^  to  a  mixed  quantity. 

Ans.  3  a  —  ^-j. 

.    ^  ^        a^  _f_  a7;  +  6\  .     ,  ^.^  ^^ 

6.  Reduce to  a  mixed  quantity. 

a  "  yt 

Ans,  a  +  b  A . 

^    ^    .        12a^  +  4a  — 3c^  •     ^  ...  ^ 

7.  Reduce , to  a  mixed  quantity. 

Ans.  2a  -{-  1  —  j-. 

.    _    ,       lOcx  +  a  —  6  ^  .     ,  ^.^  ^^ 

8.  Reduce ^ to  a  mixed  quantity. 

^^                                ^        a;     ,  «  — ^ 
Ans.  DC  H ^ — . 

n    -n  ^        a;^  +  2^2/  +  !/'  +  ^  ^  .      ,  ,.^  ^^ 

9.  Reduce —, — to  a  mixed  quantity. 

x  +  y 

X 

Ans.  X  +  y  -^ ; — 

^      x  +  y 

X^  — -  6c^cf  — —  77i 

10.  Reduce ^—z to  a  mixed  quantity 

.  X'^ 771  „ 

Alls.  —^—5 2a 

2cd 

11.  Reduce  — ^7-7 —  to  a  mixed  quantity. 

\    «>.  u. «'  +  «^' 


CASE  m. 

111.  To  reduce  any  fraction  to  the  form  of  an  entire 
quantity. 

It  is  evident  that  if  an  algebraic  quantity  be  in  the  form  of 
a  fraction,  and  the  fraction  in  its  lowest  terms,  it  will  not  re- 
duce to  an  entire  quantity  by  the  last  case.  But  the  principle 
of  negative  exponents  enables  us  to  express  the  value  of  any 
fraction  whatever  in  the  form  of  an  entire  quantity. 

1.  Reduce  ^  to  the  form  of  an  entire  quantity. 

Give  Case  III. 


REDUCTION.  83 

OPERATION.  Analysis.  It  has  been 

shown  that  a  fraction  is 

-  =  a  X  -:  =  a  X  c-'  =  ac-'         ^^"^^  *^  *^®  product  of 
O*  C^  its   numerator   into    the 

-.  -.  reciprocal  of  its  denomi- 

nator; or  ^  =  a  X  -^  (99).   But  —  =  c~^  {SO);  whence  the  ex- 
pression becomes  a  X  c~'  =  ac~^. 

From  this  example  we  deduce  tbe  following 

Rule.  Reduce  the  fraction  to  its  lowest  terms^  and  then 
r)iuUiply  its  numerator  by  the  reciprocal  of  its  denominator 
expressed  by  negative  exponents. 


examples  for  practice. 

Reduce  the  following  fractions  to  the  form   of  entire  quan- 

titles :  — 


2. 

a'b 

c'' 

Ans.  a^bc-^. 

3. 

Ans.  m^a^^b^^C^. 

4. 

Sa' 
26V 

Ans.  3  X  2-'a'b-^c-'\ 

5. 

aVcwi 
ax^cm*' 

Ans.  ax-^nr^. 

6. 

X  —  y 
X  +  y 

Ans.  {X  —  y)(x  +  y)-\ 

7. 

a^  +  2( 

ic  -f  c* 

Ans.  (a  +  c)  (a  —  c{''\ 

8. 

ahn  — 

2am^  -{-  m'' 

Ans.  mz  (a  —  m)""*. 

9. 

x^  —  ^x'z'  +  z^ 

Ans 

.  (X'  -  2O  (a;^  +  sT'P^\ 

Give  Analysis.     Rule. 


g4  FKACTIONS. 

112.  Since  a  factor  with  a  positive  exponent  may  be  trans- 
ferred from  the  denominator  to  the  numerator  by  making  its 
exponent  negative,  a  factor  with  a  negative  exponent  may  be 
transferred  by  making  its  exponent  positive.  Hence  we  have 
this  general  conclusion  : 

A  factor  may  he  transferred  from  either  term  of  afrac- 
Hon  to  the  other,  by  changing  the  sign  of  its  exponent, 

EXAMPLES  FOR  PRACTICE. 

1.  Reduce  -^7^  to  the  form  of  an  entire  quantity. 

Ans.  a^bc^. 

2.  Reduce :r—^  to  the  form  of  an  entire  quantity. 

7n~  x~  *^ 

Ans,  Soc^y^m^. 
c(a  — —  7yC\ 

3.  Reduce  —7-^^ tAt-  to  the  form  of  an  entire  quantity. 

^  ^  Ans.  c(a'  —  m'y 

4.  Reduce      _^  _^  to  the  form  of  an  entire  quantity. 

Ans.  ab'^c^, 

X  ~  ^b^  b^z^ 

5.  Reduce to  positive  exponents.  Ans.    — . 

cz-^        "^  "^  cx^ 

'fyi(x  — —  IJ)^^ 

6.  Reduce  — ^^^ ^^^—  to  positive  exponents. 

^  .  m 

Ans, 


X'  —  y' 

Reduce  the  following  fractions  to  forms  having  only  known 
quantities  in  the  numerators  and  unknown  quantities  in  the 
denominators. 

i   --  Ans    '"^"'     ■ 

c-'xz  «*z 

9    ^^  Ans    ^~' 


How  may  a  factor  be  transferred  from  one  term  of  a  fraction  to  the 
other! 


REDUCTION.  85 

CASE  IV 

113.  To  reduce  a  mixed  quantity  to  a  fraction. 

1.  Reduce  2|  to  a  fraction. 

^16        ,  ^,      16      3       16  +  3      19 

Analysis    2  =  -^ :  and  2|  =  -g-  -{-  -g-  =  — g —  =  -g-. 

X 

2.  Reduce  a  +  ?-  to  a  fraction. 

0 

OPERATION  Analysis.   It  is  evident  that  a  =« 

,x__  ah  +  X  y-    But  y  =  ah  X  y,  (99) ;  also 

^       b'^       b  X  1        ^^^.        1^^, 

—  =x  X  7" ;  and  ao  times  r-  addea 
Ob'  b 

to  X  times  r-  is  equal  to  a6  +  a;  times  7-,  or  (a6  +  a;)  X  7-,  which 

ab  +  X    ^ 
IS  equal  to  — 7 — ,  the  answer. 

The  algebraic  operation  is  exactly  like  the  arithmetical,  and 
governed  by  the  same  principle.     Hence,  the 

Rule.  Ilultiply  the  entire  jjart  by  the  denominator  of  the 
fraction  ;  add  the  numerator  if  the  sign  of  the  fraction  he 
plus,  and  subtract  it  if  the  sign  be  minuSy  and  write  the  result 
over  the  denominator . 

EXAMPLEb   FOR   PRACTICE. 

3.  Reduce  7^  and  ax  -^ —  to  fractions. 

^  A        ,a       .acx  +  b 

Ans.   %^  and . 

^  c 

X 

4.  Reduce  3  —  ^  and  x^ to  fractions. 

y                          x^ti  — —  X 
Ans.  4  and  — ^ . 

y 

5.  Reduce  y  —  1  +  = — — ^  to  a  fractional  form. 

y  +  l 

Give  Case  IV.    Analysis.    Rule. 


g5  FRACTIONS. 

6.  Reduce  .c  -f  V  H to  the  form  of  a  fraction. 

^  +  y 

x  +  y 

7.  lleduce  4  +  2x  +      to  a  fraction. 

^  ,        4:c  +  2cx+h 

Ans. . 

c 

.    ^    .        r         2x  +  6  ^     ,.  .        13x  — 5 

8.  Reduce  5a; ^ —  to  a  fraction.      Ans.  -5 — . 

6  o 

9.  Reduce  Ba  —  9 —7,-  to  a  fraction, 

a-i-  6  3 

Ans, 


a  +  3 

2ax  -fa'  ,       .  ,        (x  +  aY 

10.  Reduce  x  A to  a  fraction.     Ans. -. 

X  X 

& 

11.  Reduce  a  +  6  H — -r  to  a  fraction. 

Ans. ;— 

2a:*  ft^  4-rr* 

12.  Reduce  a  4-  cc  H — to  a  fraction.    Ans. . 

a  —  X  a  —  X 

ax  n       '  A        ^  —  2aa; 

13.  Reduce  a to  a  rraction.         Ans. . 

a  —  X  a  —  X 

CASE   V. 

114.  To  reduce  fractions  to  a  common  denominator. 

1.  Reduce  -,  -,  and  -,  to  a  common  denominator. 

x'  y  z 

Analysis.  "We  multiply  both  terms  of  each  frao 
tion  by  the  denominators  of  the  others;  that  is,  the 
terms  of  the  first  by  yz^  the  terms  of  tiie  second  by 
iC2,  a,nd  the  terms  of  the  third  by  xy.  This  process 
cannot  alter  the  values  of  the  fractions  (105,  III.,) 
and  it  must  reduce  them  to  a  common  denominator, 
because  each  new  denominator  is  necessarily  the 
product  of  all  the  given  denominators.     Hence,  the 

iA 

"What  is  Case  V  ?     Give  analysis. 


a 

ayz 

X 

b 

y 

xyz 
xbz 
xyz 

c 
z 

^xyc 
xyz 

REDUCTION.  87 

lluLE.  Multiply  each  numerator  by  all  the  denominators 
except  its  own,  for  the  new  numerators ;  and  all  the  denomi- 
nators together  for  a  common  denominator. 

Note. — Mixed  quantities  must  first  be  reduced  to  fractious,  and  entire 
quantities  to  tractioual  forms  by  writing  1  for  a  denominator. 

EXAMPLES   FOR   PRACTICE. 

8a:  26 
2.  Reduce  -^,  x-»  and  d.  to  a  common  denominator. 
2a  6c 

,        9 ex  4ab       ,  Qacd 
Ans.  ^— ,  ^-,  and  — — . 
bac  bac  bac 

8    Reduce  — -,  — ,  and  — ,  to  a  common  denominator. 
m^  mx  c  *^ 

acmx  bc'^m'^        ,   m*x 
Ans,    — r — ,  -v-,  and  —. — . 
m^xc    77VXC  nrxG 

3  2x  2x 

4.  Reduce  7-,  -^-,  and  a  +  -— ,  to  a  common  denominator. 

4  o  a 

da    Sax       ^  12a''  -4-  24x 

Ans.  ~— ,  —-^ ,  and :r- . 

12a   i2a  12a 

5.  Reduce  ,  — , — ,  and  -— - — -,  to  a  common  deno- 

.     .  ^  —  y  iXi  -\-  y  x^  -{-  y^ 

mmator.  :/      ^  i/  ^  if 

a{a^  -f  xy"^  -f  x'^y  +  y^)  m(a^  +  xy'^ —  x^y — y^)  z^x"^  —  y^) 

^ns,  7         7 , 7 , — — . 

X  —  2/*  x  —  y  X — y* 

a  "4"  c       X  a 

6.  Reduce  , ,  and  -,  to  a  common  denominator. 

X      a  —  G  b 

.        b(a^  —  c^)    bx^  .  axfa  —  c) 

Ans.   T^^- {,  -—. -,  and  y-7 (. 

bx{a  —  c)  bx{a  —  c)  bx{a  —  c) 


CASE  VL 

115.  To  reduce  fractions  to  their  least  common  de- 
nominator. 

Since  a  fraction  can  be  reduced  to  higher  terms  only  by 
multiplication,  each  of  the  higher  denominators  it  may  have 

Give  Rule.     Repeat  Case  VI. 


1^  FRACTIONS. 

must  be  some  multiple  of  its  lowest  denominator.  Hence,  a 
common  denominator  for  two  or  more  fractions  must  be  a  com- 
mon multiple  of  their  lowest  denominators,  and  the  least  com- 
mon denominator  must  be  the  least  common  multiple. 

C  771/ 

I.  Reduce  -r^  and— r;  to  their  least  common  denominator. 

ab^         a^b 

OPERATIOlSr. 

a^b^  ~-  ab^  =  a]  and  c  X  a  =  ao, 
a^b^  -r-  a^b  ==b]  and  m  X  6  =  6m. 
c         ac   ^  m        bm 

ab'  ^~aW  '  a^  ^'¥^'' 

Analysis.  We  find  by  inspection  that  a^b^  is  the  least  common  mul- 
tiple of  the  given  denominators ;  it  is,  therefore,  the  least  denomi- 
nator to  which  the  fractions  can  be  reduced.  To  ascertain  what 
factor  will  reduce  each  denominator  to  a^i^,  we  divide  this  term  by 
each  denominator,  and  obtain  a  and  b.  Since  the  given  deno- 
minators must  be  multiplied  by  a  and  b  respectively  to  reduce  them 
to  the  required  denominator,  the  corresponding  numerators  must 
also  be  multiplied  by  these  factors  for  the  new  numerators ;  and  we 
have  c  X  a,  or  ac^  for  the  first  numerator,  and  m  X  b,  or  biUj  for  the 

second  numerator,  and  — —  and  — r,  the  answer. 
a^b  d'b 

Hence  the  following 

Rule.  I.  Find  the  least  common  multiple  of  all  the  denO" 
minators  for  the  least  common  denominator. 

II.  Divide  this  common  denominator  by  each  of  the  given 
denominators,  and  multiply  each  numerator  by  the  corre- 
sponding quotient.  The  products  will  be  the  new  nur 
merators, 

EXAMPLES   FOR   PRACTICE. 

Reduce  the  following  fractions  and  mixed  quantities  to  then 
least  common  denominator. 

*     9    ^    ^    or,^     ^  A        h^"cdm    acdx        _    ahh 

2-  ac'  ¥c'  ^""^  ^  ^"«-  abVd'  ^d^d'  ^"'^  a^d- 

Give  analyBis.     Kule. 


REDUCTION.  89 

«    m  c  4-  m        ^  d  .       mbc  dbc  +  abm       ,   acd 

3.  -;„ ,  and  —:,      Arts.  -,7-, 57 1  and  -^r-. 

a^'     ac  ah  a^hc        a'bo  a^bo 

A   ^  +  ^  g  —  b  a^ 

iacx^  -f-  4:bcx^  ^o?c  —  6a&(?       ^     3a*a? 
^ns.  ...... ,  — T7T-0— 1 — ,  and 


^    e  —  d  X        ,  .        ac  —  ad    bx        _  a^6m 

5.  — r— ,  -2,  and  m.  Ans.  :pr—,  -^,  and  — ^. 

ah      a^  a^b        a^b            a^b 

or     o  aocu  ■  I  ■  cc/      c            oc  ij 

6.  a  +  -,  — ,  and  ic.  Ans.  —^ ,  — ,  aiid  -^. 

y  xy  xy         xy            xy 

^        a            b            1  <^ 

7.  ,  — j — ,  and 


x  —  y'  X  '\'  y'  x^  —  y^ 


a(x  +  y)     bfx  —  y)         ,        c 


x'^  —  y^  *    x^  —  2/*  '  x^  —  y** 


8,       ^ 


^ITi'Sr^T'^'^'i^rzn- 


.        x^  4-  x^  A-  x^  +  X  X*  -{-  x"^        _       ic* 
Ane.  ,— 3 .  -^—^ ,  and  ^— ^. 


X* 


^    a  —  b       ,      a  —  b 
9.  and 


ao  a(a  +  6)* 

a'  —  ¥  T     cfa  —  5") 

Ans,  — — yr  and  —^^ — — r^. 

ac(a  +  0)  ac(a  -f-  6) 


10.  .^^-^  and 


i(l  —  m^)  c(  1 —  m)  * 

j_ng  ?£ and  "^^^  "^  ""^ 


8* 


00  FBACTIONS. 


ADDITION. 

fil6.  We  have  seen  (50),  that  entire  quantities  may  be 
added  wlien  they  have  a  common  factor,  to  serve  as  the  unit 
of  addition.  In  like  manner,  fractions  may  be  added  when 
they  have  a  common  unit;  and  since  the  fractional  unit  is  the 
reciprDcal  of  the  denominator,  fractions  to  be  added  must  have 
a  common  denominator. 

CL  C 

1    What  is  the  sum  of  ^  and  -  ? 
h  0 

Analysts.     The  fractions  have  a  com- 

OPERATION.  I  ^ 

d        Q        a  4-  c         ^^^  ^^^*'  tT*    ^^  T  *^^^^  ^^^^  ^^  taken  a 

times,  and  in  j-  it  is  taken  c  times ;  hence, 
in  the  sum  of  the  fractions,  it  must  be  taken  a  plus  c  times,  expressed 

2.  What  is  the  sum  of  -,  — ,  and  -, —  ? 
0    on  bm 

OPERATION. 

J-  =  J \  Analysis.     We  first  reduce 

/„      .  ,      ,       the    given    fractions    to   their 

n  rrti    f  Fractions  reduced  ,          "                           .             . 

_  =  ±11.  V     to  a  common  de-  ^^^     commoD     denominator, 

671  bmnf    nominator.  and   then   add  as  in  the  first 

d  dn  \  example. 

bm  bnin' 

amn        cm         dn   amn  +  cm  -f  dn 

bmn  bmn       bmn                 bmn 

From  these  examples  we  derive  the  following 

Rule.  I.  Reduce  the  fractions  to  their  lead  common  de^ 
nominator. 

II.  Add  the  numerators^  and  write  the  result  over  the 
common  denominator. 

What  is  a  fractional  unit.    Give  analysis  of  Addition  of  Fractions.     Biile. 


ADDITION.  91 


Notes.  1.  When  there  are  mixed  quantities,  the  entire  quantities  and 
the  fractions  may  be  added  separately ;  or  the  mixed  quantities  may  be 
1  educed  to  fractions  and  added. 

2.  A  tractional  result  should  be  reduced  to  its  lowest  terms. 


EXAMPLES   F<)R   PRACTICE. 

3.  Wliat  is  the  sum  of  .-^  and  — :, —  ? 


4.  What  is  the  sum  of  -,  — ,  and  --  ? 
y    axi^         X 

Ans. 


ba  +  Sm 

Ans. 

66 

1  a 

+  z  +  y' 

xy 

5.  What  is  the  sum  of  -  and ? 


he 

X 


6.  What  is  the  sum  of  ^,  q,  and  j  ?         Ans.  x  +  y^. 

Z        O  4:  Jim* 

rj^  2  \X 

7.  What  is  the  sum  of  — ^, —  and  ~  1  ^Oo: 14 

'       J„s.  — 2j— 

11  2a 

8.  What  is  the  sum  of r  and -r  ?    Ans.  -f 


a  -^  b         a  —  b  '  d^ — 6** 

9.  Add  -^-  to  -^—,  Ans.  ^l+y]. 

X  -j-  y       X  —  y  x^  —  2/ 

10.  Add  — -^-^ to  — ;:: Ans.  — = . 

SOG               ic  bo 

11.  Add  :j — j — ,  = ,  and  = — ; — .  Ans, 


1  +  a'  1  —  a'         1  +  a  1  — a  ' 

12.  Add  — ,  — ,  and  j-.  Ans,  r-^-^ . 

b    ob  ia  I2ab 

^^^  6a6  — 86»— .12acf  166c      ^3a  — 46 

13.  Add =-c|- and  — ^r—  . 

126e  36 

2a  — 6 

-4  ns.  — -2 . 

4c 


I  FRACTIONS. 

14.  Add  2a:,  Sx  +  x»  ^^^  ^  "^  T"        ^^'  ^^  "*"  "45 


15.  Add  5a;  +    '  ^      s-i^d  4a7  + 


3      **""•"  ^      5^    • 

^ns.  9x  -A '- — =-^ 

15a: 


^h                              1 
16.   Add TTT— T-Tx    a^^  — ;-?• 

a  —  b 

17.  Add  — r— ,  — r — ,  and . 

ab         be                ac 

Ans,  0. 

18.  Add and . 

.        a  —  X 
Ans, , 

X 

19.  Add  — ^--, ,  and  5-.      ^ns.  o  - 

y  ay  da  Say 

^.,     ,  ^,a  +  b  ^    a  —  b  .        2(a^  +  b^) 

20.  Add  — "-T  to  — — T.  Ans.     \       rr^, 

.  ,  -  a  a  —  86        ^  a'  —  5*  —  ab 

21.  Add  -7, J—,  and r--^ . 

^       ^^  ^^'^    ,        acd^W  +  a' 

Ans.  J-— . 

bed 


22.  Find  the  sum  of r  and 7.        Ans.  —, 7- 

a  -i-  b         a  —  b  a^  —  b' 


,2* 


1                          V  X 

23.  Find  the  sum  of and    „        _       Ayis.  -^ -,. 

X  +  y         x^  —  y^  x^  —  y^ 

24  Fmd  the  sum  of  = r  and  v—; — s*       ^ws.  ^ ^ 

1  —  a*         1  -f  a^  1  —  a^ 

25  Find  the  sum  of  — hi-  and  1  —  ( — -, — \ 

a      b                  \    ab    J  Ans,  L 


SUBTRACTION.  gg 


SUBTRACTION. 

117,  We  have  seen  (54),  that  one  entire  quantity  may  be 
subtracted  from  another,  when  they  have  a  common  factor  to 
serve  as  the  unit  of  subtraction.  In  like  mannei,  one  fraction 
may  be  subtracted  from  another  w^hen  they  have  the  .^me  frac- 
tional unit,  or  a  common  denominator. 

I.  From  -'  subtract  ^. 

0  0 

Analysis.    The  fractions  have  a  com- 
OPERATION.  2  a 

d        Q        d ^         mon  unit,  -r-     ^^  Tt  this  unit  is  taken  a 

times,  and  in  -j-  it  is   taken  c  times ; 

hence,  in  the  difference  of  the  fractions,  it  is  taken  a  minus  c  times, 

a  —  c 
expressed  — r — . 

From  this  example  we  derive  the  following 

Rule.  I.  Reduce  the  fractions  to  their  least  common  de^ 
nominator, 

II.  Subtract  the  numerator  of  the  subtrahend  from  the  nu- 
merator of  the  minuend,  and  write  the  r^esult  over  the  common 
denominator. 

Note. — Mixed  quantities  must  be  reduced  to  fractions  before  subtract^ 
ing;  and  fractional  results  should  be  reduced  to  their  lowest  terms 


EXAMPLES   FOR  PRACTICE. 

*jx  2x 1 

2  From  -r-  take  — ^ — . 

21x  —  4:x+2       17rr-f2 

.^nS,  yi ■ =  7^ • 


Un^er  what  circunjstance^-  can  one  fraction  be  subtracted  from  an- 
other t    (Stiyia  an&tysis.    Rule. 


94  FRACTIONS, 

3.  From take 


x^y  x-\-y 

OPERATION. 

1  1        ^  +  2/      ^  —  y ^y 


X  —  y       X  -\-  y       x^  —  y^  x'^  —  i/^       ^^  —  2/** 

4.  From  .^  take  -~.  Ans.  ^ 

_     _         2ax      .     K>ax  .              Wax 

5.  From  -^  take  -^.  An%, ^r-. 

8               2  b 

-I                                         ^^  C}  q 

6.  From  — ; — --  take  -^ ■=*  An%, 


a  +  1  a=^  —  a  +  1  '   1  4.  a'*' 

.    _  3^.5  .        3.T  -~  10a 

7.  From  -r-  take  ,7-.  -ans.  -. 

4a  Ix  4ax 

8.  From  7-  take  rr-,  ^ns,  — =-x . 

4:X  oa  iZax 

I  2  3 X 

9.  From ^  take  — — ^.  ^ns.  -^ ='. 

X  —  1  X  +  1  x^ —  1 

10.  From  2a  —  2x  +  ^      ^  take  2a  —  4a;  +  ^       ^. 
a  a; 

a'^  —  a;' 


Ans.  2x  + 


ax 


11.  From  ^^ take  — ;;; .  Ans.  -^^^-r? . 

oc  ic  35c 

,„    _        5x  +  l,i     210^  +  3  ^  127^+17 

12.  From  — - —  take .        Ans. ^1 . 

7  4  ^o 

13.  From  --^—^  take     ^    ^.  -4ns.   — ;;-^-  ^. 

-:a  oa  6a 

1.1     T7         1  +a\  ,     l  —  a^  .  4a' 

14.  From  = ^  take  zr— - — j.  Am.  = ^,. 

1  —  a^  1  4-  tt*^  1  —  a 

15.  From  x  +  ^'pL  take  -^-^. 

X—  '^y 


SUBTRACTION.  95 

16    FrQm  — ^ —  take  — ^- — . 

17.  From  ^^ — ^s-^  take  ,.  -472s.  ^, 

a  —  0  a  -|-  0  a  —  o 

18.  From ^  take ,  Ana, 


X  —  3  X     *  '  x^  —  3x* 

19.  From  6a  H —- —  take  4a  +       T"  • 

Ans.  2a  +  ^ti. 

0 

20.  What  is  the  value  of  ^^^^~ ^  ? 

a^  —  ^2        ^  —  ^  ^ 

^WS,    — r-T. 
a  -f  O 

21.  What  is  the  value  of  l-±^*      Irifi'  f 

l  —  x^       I  +  x^  4^ 


22.  What  is  the  value  of 


1— X*' 
a  —  X       a*  —  x^  „ 


Ans. 

X 


23.  What  is  the  value  of  ^^  +  -=^  —  ^-  ? 

oc  ac  ab  o 

a 

rt'^    -rrri    .  .     1         i        ^  3x       2x       5.T  _  ,        21a; 

24.  What  IS  the  value  K)f  -7-  +  -= ^  r         ^ns.   -77^. 

4         5         o  40 

25.  From take v  ^^^^  —^ — — 

n  n  —  1  n  —  n 

26.  From  -z r,  take  -z .  Ans, 


1  —  x'^  1  —  X  '   1  -f-  x' 

27    From         4+^—   take  , ^i^— r.. 

(rt  —  b)  {x  —  a)  {ci  —  6)  (x  —  6) 

-,  x  -4-  c 

,  (^a:  —  a)  (x  —  0) 


96  FRACTIONS. 

MULTIPLICATION. 
CASE  I. 

118.  To  multiply  a  fraction  by  an  entire  quantity, 

1,  Multiply  -by  c. 

Analysis.     A  fraction  may  be  multiplied 
OPERATION.  ^^  multiplying  its  numerator  (105,  I)  ;    ^N-e 

£j  dQ         therefore  multiply  the  numerator,  a,   by  c, 

0  o  and  obtain  for  the  required  product,  -r-- 


2.  Multiply  —  hj  X, 

Analysis.    A  fraction  may  be  multiplied 

OPERATION.  ^y  dividing  its  denominator  (105,  11) ;  we 

therefore  divide  the  denominator,  xy,  by  x, 

mm  ^ 

-— -  X  ^  =  —         and  obtain  for  the  required  product,   — . 

*^y  y  y 

Hence, 

Multiphjing  a  fraction  consists  in  multiplying  its  numera- 
tor, or  dividing  its  denominator. 

EXAMPLES   FOR  PRACTICE. 

c  -  .        cm  . 


3.  Multiply  J  by  m.  Ans,    -r-, 

a'^x 

4,  Multiply -j  by  ax.  Ans, 


c  —  df 

5.  Multiply  -j-z  by  cd.  Am,  — . 

a   Multiply  g|  by  7.  V  Ans.  -^• 

7    Multiply zrhj  a  +  X.  Ans.  r-. 

X  — —  X  X   — —  X 

8.  Multiply  =- — - — -  hj  X  +  y.  Ans.  -r- 

^''  h(^x  +  y)    -"          ^  b 

What  is  Case  I  ?      Give  analysis.  Deduction. 


MCJLTIPLICATIOK.  97 

,.  9.  Multiply  —:, 5  by  m  —  v.  Ans.  -— 

I  ^  ^  m'  —  y^    '  "^  rn  +  y 

10.  Multiply  — by  x^  +  !•  Ans.  --g . 

iu    ■       X  Sj    —•—  X 

119.  It  is  usually  advantageous  to  indicate  the  multiplicap 
tion,  and  apply  cancellation  before  obtaining  the  actual 
product. 

1.  Multiply  -  by  x. 

OPERATION.  Analysis.    Having  indicated  t1i«  mul- 

tiplication, we  cancel  the  common  factor, 
^  ^  ^  _.  ^  a;,  from  both  numerator  and  denominator, 

X  and  we  have  a  for  the  product. 

2.  Multiply  ^  by  6m. 

OPiCRATlON.  Analysis.    Having  indicated  the  mul- 

c  X  6m  tiplication,  we  cancel  3m,  and  obtain  2c 

— g— —  =  2o  for  the  product.    Hence, 

1.  A  fraction  is  multiplied  by  its  own  denominator  by 
simply  suppressing  the  denominator. 

11.  If  a  fraction  be  multiplied  by  its  own  denominator,  or 
by  any  multiple  of  that  denominator,  the  product  will  be  an 
entire  quantity. 


^ 


EXAMPLES  FOR  PRAOTIOB. 


35 

3.  Multiply  —  by  t/.  Ans,  x. 

4.  Multiply  -Tj-  by  5ft.  Ans.  Sax. 

cd} 

5.  Multiply by  a  —  x.  Ans.  cd}. 

a     '  X 

When   may  cancellation  be  applied  in  mnltiplioatlon  ?     Give   fixvt 
dedutrtioi^.    Seodnd. 

9  O 


98  FRACTIONS. 

6.  Multiply  —  by  2x».  Ans.  6^V. 

cc 

7.  Multiply  ~~7r-^  by  20.  Ans.  6a  —  2x. 


10 

a  —  X  . 


8.  Multiply  -ggg-  by  896. 


-4ns.  a  —  x. 


9.  Multiply  -2 3  by  a  +  a;,  Ans, 

10.  Multiply  -^  by  x^  —  1.  Ans.  3c(a:  +  1). 

1 1.  Multiply  ^5-i^  by  a"  —  2ab  +  h\  Ans.  a"  —  h\ 


CASE  n. 

120.  To  multiply  an  entire  or  a  fractional  quantity 
l)y  a  fraction. 

1.  Multiply  a  by  -. 

FIRST  OPERATION.  ANALYSIS.    It  IS  evident    that    the 

product  of  two  quantities  is  the  same, 
^  X_-__  y^  as=^         whichever  be  taken  as  the  multiplier ; 
COG  h 

consequently  a,   multiplied  by  — ,  is 

eqaal  to  --  multiplied  by  a ;  and,  according  to  Case  I,  —   multiplied 

.  «& 
by  a,  IS  — . 
c 

SECOND  OPERATION. 

Analysis.  "We  first  reduce  the  multi- 
6         ,     ,  b 

—  =00  plier,  — ,  to  an  entire  form,  5<r*  by  (111). 

7}  ah        Then  a,  multiplied  by  bcr^  is  dbc^, 

rt  X  -  =  ahcr^    =  —  ^5 

^  ^         which  is  equal  to  — ,  (112),  as  before, 

Gi?v^  Case  II.     Analyses. 


MULTIPLICATION.  99 


CI  c 

2.  Multiply  -  by  — . 


OPERATION. 


a        c  .  .  .      .       ac 

—  X  —  =  (ix~^  X  cwr^  =  acar^mr^  =  — . 
X       m  xm 

Analysis.  By  (111),  —  is  equal  to  oa;"*,  and  —  is  equal  to  cm'^ ; 

iind  ax~^,  multiplied  by  cm~\  is  acx'~^m~^f  which  is  equal  to  —  (112). 

By  inspecting  this  result  we  perceive  that  the  numerator,  ac,  is  the 
product  of  the  given  numerators,  and  the  denominator,  xm,  is  the 
product  of  the  given  denominators.     Hence,  the 

Rule.  I.  Reduce  entire  and  mixed  quantities  to  frac' 
tional  forms. 

II.  Multiply  the  numerators  together  for  a  new  numera- 
tor, and  the  denominators,  for  a  new  denominator,  canceling 
all  factors  common  to  the  numerator  and  denominator  of  the 
indicated  product. 

EXAMPLES  FOR  PRACTICE. 

o    HIT  IX-  ,    Ba ,     5a;  .        16a 

3.  Multiply  j^  by -g.  ^^«-    32' 

4.  Multiply  I  by  |.  ^ns.  £. 

5.  Multiply  Ig  by  g.  Ans.  ^ 

6.  Multiply  ^A-  by  -.  Ans.  ^J^±^. 

c  z  cz 

a  —  h  25a;  — 25        ,      1      ,       ,, 

7.  Multiply  — F — ,  — 2 TJ-,  and =-  together. 

OPERATION. 

a  —  h  2b{x  —  l)       ^      1      _      5 


(a +  6)  (a  —  6)      x  —  1       a  +  b 


Give  ankTysis.    Rule 


100  FRACTIONS. 

Analysis.  We  resolve  the  quantities  into  convenient  factors, 
and  indicate  the  multiplication.  Since  5,  a  —  6,  and  x  —  1,  are 
factors  common  to  the  numerator  and  denominator  of  the  indicated 
product,  we  suppress  or  cancel  them,  and   obtain   for  the  result, 

8    Multiply ,  — '^ — ,  and together. 

Arts.  — , 

9.  Multiply  ^  by  -T-. 

10.  Multiply  -g-  by  -^. 

-11     Tir  7x.  1    2aj ,     Zab      Sac 

11.  Multiply -by- X^- 

12.  Multiply  — -by— ^. 

13.  Multiply  -,  — ,  ^  together. 

14.  Multiply  (^-^)  by  ^. 

15.  Multiply  -^^  by  g^.  Arts.  ^^^ 

4:(7QC     QX1J  2 

16.  What  is  the  product  of ,  -^-,  and  —  ?    Ans,  12a;. 

^  y      Za  X 

17.  What  is  the  product  of  ^rr- —  into  — '-^^ —  ? 

'^  36  +  c  oah 

4ac — 2bc 
^"^^  TEb^irSbG 

18.  Multiply  5  +  ^  by  ?.  Ans.  ±t^. 

a    '  as  X 

19.  Multiply  -g^  by  -^-^.  Am.  ^.—^ 


Ana. 

dax 
2b' 

Ans. 

a^x. 

Ans.  9ax, 

Ans. 

ay 

XI 

Ans. 

4a 
2* 

Ans. 

a 

18' 

2x  +  Sy 

MULTIPLICATION.  101 

^2 ^i  2a  J,         (« —  ^)^ 

20.  Multiply  -2~  by  — --.  Ans.  — ^ 

21.  Multiply  ^.  -}-^,  and  ^.  ^n«.  a. 

22.  Multiply  3a,  ^,  and  J=^  together.    ^^^^_  ^^ 

^"®-  ^(^  6)' 

3_j,» 535  7a  .        Saa;—  5a 

23.  Multiply  —14—  by  2^_3a;-        ^"*-  ll^^  6"' 

3cB»      ,     15x  — 30  .        9a; 

24.  Multiply  g^^^jo  by  —2^-  ^na.  -^ . 

25.  Multiply  ^-"-^y  4.  ^ns.1. 

26.  Multiply  —--^  by  j^.  -*7i«.  ^p-. 

2T.Mult.ply^by^±L^. 

28.  Multiply  ^^  by  ^-j-^  ^"8-  -;;fzr^ 

(a  —  xf  3rt6      „       2c       , 

29.  What  is  the  value  of  —1^-  X  -„-Z:i-  ^  (a  —  x)'  ' 

.4ns.  36e. 

30.  What  is  the  value  of  -^^^^  X  ^^p^^e  r      ^  ^ 


.  «.  -r  6 


9* 


102  FRACTIONS. 

DIVISION. 
CASE  I. 

121.  To  divide  a  fraction  by  an  entire  quantity. 

CLCC 

1.  Divide  -^  by  x. 


Analysts.     A  fraction   may  "be  divided 
OPERATION.  by    dividing    its    numerator     ( 105,    I ) ; 

we   therefore   divide  the  given  numerator, 
aXf  by  Xy  and  obtain  for  the  required  quo- 
a 


ax  a 


tient,  \, 


2.  Divide  —  by  a. 


ANALYSIS.     A  fraction  may  be  divided 

OPERATION.  by    multiplying    its     denominator     (105, 

II) ;  we  therefore  multiply  the  denomina- 

r-  C  =  —  tor,  c,  by  a,  and   obtain  for  the  required 

C  OG  ^ 

quotient  — .    Hence, 

Dividing  a  fraction  consists  in  dividing  its  numerator  or 
multiplying  its  denominator, 

EXAMl^LES   FOR   PRACTICE. 

3.  Divide  — ^  by  Sax,  Ans.  -.. 

cd  ca 


r 


4.  Divide  -i— ^  by  Zb\  Ans.   -. — r 

5.  Divide  — ^—  by  2m.  Ans,  -?,——. 

m.z  Znvz 

6.  Divide by  y,  Ans, :^ 

x  —  y       -^  xy—if 

7.  Divide  '^~~  by  a  +  1.  Ans.  -^— . 

cd  cd 

8.  Divide  -2 T  by  a"^  +  4.  Ans. :r^. 

a^  —  4    "^  c*  — 16 


What  is  Case  I?     Give  analyses.     Deduction. 


DIVISION.  •  103 

CASE  n. 

l^a  To  divide  an  entire  or  a  fractional  quantity  by 
a  fraction. 

1.  Divide  m  by  -. 

•'  X 

Analysis.    We  first 
OPERATION.  ^^^^^^   ^^^   ^.^i3^^^   a^ 

X 

^  —  n^-i  to  an  entire  form,  ax"^. 

—  tt»C/       « 

X  Then  m  divided  by  occ"* 

is  _^,  which  is  equal 


a             .       ,-1         ^          ^'^^ 
X             *                 aa3~^         a  * 

ax-' 
to  7,  (112), 

2.  Divide   ^-  by  J. 

OPERATION. 

6       d 

Analysis.     -  is  equal  to  ab'-'^f  and  -  is  equal  to  cd"^ ;  and  a5~*  di- 
0  d 

vidod  by  cd~^  is  -  _,  which  is  equal  to  -— ,  (112).    By  inspecting  this 
cd  be 

result,  we  perceive  that  the  numerator,  ad,  is  the  numerator  of  the 
dividend  multiplied  by  the  denominator  of  the  divisor;  and  the  de- 
nominator, be,  is  the  denominator 
SECOND    OPERATION.  of  the  dividend  multiplied  by  the 

numerator  of  the   divisor.     The 
^_j_?._  ^'_y^  ^__  ^  same  result  can  be  more  readily 

b        d       b        c        be  obtained  by  inverting  the  divisor, 

as  in  the  second  operation,  and 
then  multipl3-ing  the  upper  and  the  lower  terms  together.  Hence, 
the  following 

Rule.     I.   Reduce  entire  and  mixed  quantities  to  frav- 
tional  forms. 

II.  Invert  the  terms  of  the  divisor,  and  proceed  as  in  mul 
U'plication. 

What  is  Case  II.?     Give  analyses.     Rule. 


104 


FRACTIONS. 


EXAMPLES  FOR  PRACTICE. 


_     _ .  .  _         a      .      a 

8    Divide  r by  f-. 

1  —  a       5 

.     ^.  ...    2x .     i^xy 
4.  Divide  -r  by  -—■• 
ab        ab 

5    Divide  5 —  by  — .. 

6,  Divide by  -^ ; 


7.  Divide  m  H —  by  -7. 

y        d 

8.  Divide  a  H by 

X 


9.  Divide 


2ax  +  a?^  ,  a? 


by 


a  —  X 


-.    „  .,    14x  — 3,     10a;  — 4 

10.  Dmde g—  by  — ^^ 

11.  Divide ^ —  by  -^. 


12.  Divide 


%x 


13.  Divide  -->—  by  =^. 

14.  Divide  — r —  by 


42/ 


15.  Divide  ~g-  by  -^. 

16.  Divide^  by  |5. 


17.  Divide 


a;      ,     a; 

:ri^y2' 


-4ns 


u4ns. 


.4ns. 
Ans. 


l  —  a 
2 

am 

Wz 

2c       * 
d{my  +  a?) 


Ans, 


Ans,  X  -{ . 

c 

2a  +  X 
70a;  — 15 


Ans, 

Ans, 
Ans, 


10a;— T* 
9a;  — 3 
x 
18a;  — 21 


a;=»  — 1 
Ans,  12a. 
8?/ 


18.  Divide 5^J-£.  by  ^ 


ab 


be 


Ans, 
Ans. 
Ans, 
Ans, 
Ans, 


5' 

7 

2.i' 
a  -f-  1 

2 

a;—  I 
ox  —  ,  y 


DIVISION.  ;i05 

19.  Divide  !^— -  by  ^^5-+?.  Ans,  2m  —  2n. 

20.  Divide  -r-  ^7  jst-  Ans,   -,—. 

o         do  2a 

21.  Divide  -77— f-  by  -7-^.  -dns.  -tt-;; — . 

22.  Divide    ,   ^~\,,  by  ?^^.  ^n8.  a:  +  ~ 

23.  Divide  l  +  \\)yl  —  \  Arts, 


x  —  h 

Scd 

x' 

—  6* 

a  a^  'a  —  1* 

2^.  Divide  (,-i-  +  ^-)  by  P^Z- 

\i  +  X     1 — xJ  ^  (i—xy 


Ans.  :^ ■. 

1  4-  .^;' 

6  +  1 


25.  Divide  (^  +  2.^)  by  (6  + J--1).        ^n.     ^^,  . 

123.  The  division  of  one  fraction  by  another,  or  of  one 
mixed  quantity  by  another,  may  be  indicated  in  the  form  of  a 
complex  fraction,  and  the  result  reduced  to  a  simple  fraction, 

1.  Divide  a  -| —  by  x  ■{-  —^ 
c  z 

Analysis.     We  indicate  the  division 

OPEBATION.  |3y  writing  the  dividend  above  a  horizon- 

a  '^~~-  *^^  ^i°6>  ^^^  the  divisor  below.     Then, 

C^ gC2:  +  bz  since  the  denominator  of  a  fraction  will 

y       cxz  +  cy         disappear  when  the  fraction  is  multiplied 

^    •  ^  by  any    multiple    of   its    denominator 

(119,  II),  we  multiply  both  numerator 

and  denominator  of  the  complex  fraction  by  cZy  the  least  common 

multiple  of  the  denominators  of  the  fractional  parts,  and  obtain  tho 

acz  -}-  bz 
simple  fraction, jT — .     Hence,  to  simplify  a  complex  fraction, 

Multiply  both  numerator  and  denominator  by  the  least  com- 
mon multiple  of  the  denominators  of  the  fractional  parts. 

Explain  the  process  of  reducing  a  complex  to  a  simple  fraction.  Give 
deduction. 


lOf  FRACTIONS. 


EXAMPLES   FOR  PRACTICE. 


2^ 

2.  Reduce  ^  to  a  simple  fraction.  Atis.  p. 

3.  Reduce  ^   '    "  to  a  simple  fraction.  _  _ 

b  —  i  Ans.  ^±l'Il^ 

bdn  —  en 

4.  Reduce  — 7—^  to  a  simple  fraction.  Ans.  ^- 

0  60 

-  +  c  a  4-  Ac 

5.  Reduce  - — ~-  to  a  simple  fraction.      Ans. ~. 

x  +  ^  ^  4ar  +  2z 

6.  Reduce p  to  a  simple  fraction.        Ans, 


a  +  T  ac  +  1* 

7.  Simplify  the  fraction  -,  Ans, 


1  +  ^  »i  +  wi 


8.   Simplify  the  fraction       ^       " 


1-^  +  ^ 


1 

*" .  {a  —  3>r^ 

a{x^  —  X  +1) 


9.  Simplify  the  fraction  ^Je:'     Ans.  l^^l-^ZZ^. 

^^    ■_!  n^ 

10.  Simplify  the  fraction  ^"-^^ .  Ans.  — . 

1 1.  Reduce    ^*    to  a  simple  fraction.  x^jo^  —  x) 


Jjjj^^Y" 


EQUATIONS.   K        ^  ■'^'  \\^.       107 


LA.,.-- 


ft    U 

SECTION  II.  J 

EQUATIONS.  \ 

IS4.  An  Equation  is  an  expression  of  equality  between  \ 

two   quantities;    thus,   «:  =  4,    5u:  =  60,   3x  =  a  +  &,   are  ! 

equations.  j 

Tlis  First  Member  of  an  equation  is  the  quantity  on  the  left  i 

of  the  sign  ;  and  I 

The  Second  Member  is  the  quantity  on  the  right  of  the  j 

sign  ;   thus,  in  the  equation,  a  -j-  b  =  7x  —  y,  the   quantity,  \ 

a  +  &,  is  the  first  member  of  the  equation,  and  the  quantity,  j 

7x  —  ?/,  is  the  second  member.  1 

t^fei.  An  Arithmetical  Equation  is  ope  which  expresses  | 

the   equality  of  numbers  or  sets  of  numbers ;    as  10  =  10 ;  \  | 

4+3  =  6  +  1.                                             '.  Y 

I^H.  An  Algebraic  Equation  is  one- which  contains  one  or  ] 
more  literal  quantities  ;'  as,  ox  =  Ig  ;  c{a  +  b)  =  d.  Alge- 
braic equations  serve  to  express  the  relation^  between  known  I 
and  unknown  quantities,  and  to  determine  the  values  of  the  ] 
unknown  quantities  by  comparing  them  with  some  that  are  j 
known.  / 

iSI7.  A  Numeral  Equation  is  one  in  which  all  the  known  J 

quantities  are  expressed  by  numbers ;  as,  ox  -{-  2x  =  25.  .     ] 

1^8.  A  Literal  Equation  is  one^in  which  some  or  all  the  l 

known  quantities  are  expressed  by  letters  ;  as,  x  +  3ca:  =  m;  | 
^2/  +  cc^  =  91.                                                   ^                            ^     1 

ISO.  An  Identical  Equation  is  one  in  which  the  tw^-roem^'^    hS| 

Define  an  equation.  The  members.  An  arithmetical  equation.  An 
Blgebraic  equation.  A  numeral  equation.  A  literal  equation.  An  ideu* 
tical  equation. 


108  SIMPLE  EQUATIONS. 

bers  are  the  same,  or  capable  of  being  reduced  to  the  same  ex- 
pression by  performing  the  operations  indicated.     Thus, 

9^ 1     _  2x 1") 

"  '  y  are  identical  equations, 

ox  -j-  ox  =  ox         J 

130.  The  Degree  of  an  equation  is  denoted  by  the  highest 
exponent  of  the  unknown  quantity  in  the  equation.     Thus, 

X  —  ar  ^^^  equations  of  the  first  degree. 

CC  ~y-  OX  =  C  J 

'  '^     [  are  equations  of  the  second  degree. 
ax^  -{-  Ox  =h) 

are  equations  of  the  third  degree,  &c. 


x^  =  a\ 
-{-  bx=  c) 


131.  A  Simple  equation  is  an  equation  of  the  first  degree. 

13S.  A  ftnadratic  equation  is  an  equaticto  of  the  second 
degree.  \  v 

133.  A  Cubic  equation  is  an  equation  of  the  third  degree. 

TRANSFORMATION  OF  EQUATIONS. 

134.  The  Transformation  of  an  equation  is  the  process  of 
changing  its  form  without  destroying  the  equality  of  its  mem- 
bers. ■  f 

Since  an  equation  is  only  an  expression  of  equality  l^etween 
two  quantities,  all  the  changes  that  can  be  made  in  the  mem- 
bers of  an  equation,  by  which  their  values  are  altered  without 
destroying  their  equality,  are  embraced  in  the  axioms  (4©), 
and  may  be  stated  as  follows  :  — 

I.  The  same  or  equal  quojitiiies  may  be  added  to  both 
members,     (Ax.  1.) 

II.  The  same  or  equal  quantities  may  bet  subtracted  from 
both  members,     (Ax.  2.) 

III.  Both  members  may  be  multiplied  by  the  same  or  equal 
quantities.     (Ax.  8.) 

Define  the  degree  of  an  equation.  A  simple  equation.  A  quadratie 
equation.  A  cubic  equation*  What  is  the  trauBformation  of  an  equation? 
State  the  principles  upon  which  all  transformations  are  based. 


ONE   UNKNOWN   QUANTITY.  1Q(J 

LY.  Both  membei's  may  he  divided  by  the  same  or  equal 
quantities.     (Ax.  4.) 

Y.  Both  members  may  be  raised,  by  involution,  to  the  same 
power.     (Ax.  8.) 

YI.  Both  members  may  be  reduced,  by  evolution,  to  the 
same  root.     (Ax.  9.) 

Note. — As  the  principal  object  in  transforming  an  equation  is  to  find 
the  value  of  the  unknown  quantity,  we  present  here  only  those  cases 
necessary  to  the  solution  of  Simple  Equations. 

I 

CASE   I. 

130.  To  transpose  any  term  of  an  equation.      ^ 

Traijsposition  is  tire  process  of  changing  a  term  from  one 
member  of  an  equation  to  the  other,  wit^iout  destroying  the 
equality.  \ 

1.  In  X  +  a  =  6,  transpose  a  to  the  second  member. 

OPERATION.  Analysis.     Since  the  equality     « 

,       7  of  the  members  is  not  destroyed 

by  taking  the  same  quantity  from 
both  (134, 11),  we  subtract  a  from  ^ 
X  =  b  —  a         each  member,  and  obtain  for  a  re- 
H^ult,  x  =  b  —  a. 

The  same  result  may  be  ob- 
tained by  dropping  +  «  from  the 


Subtract,  a  =  a 


Or,  x  +  a=b  . 

cc  =  6  — 

#  first  member,  and  writing  —  a  inO# 

the  second  member,  as  in  the  second  operation. 

2.  Inb  =  X  —  c,  transpose  c  to  the  first  member. 

OPERATION.  Analysis.    We  add  c  to  both 

b  =  X  —  c  members  of  the   equation  (134, 

Add,           c  =  c  I)»  and  obtain  b  -\-  c  =  x.     The 

b  4-  G~^x  same  result  may  be  obtained  by 

dropping  —  c  from    the    second 

Or,            b  =  x  —  c  member  and  writing  +  c  in  the 

b  -\-  c  =:  X  ^^^^  member. 

What  is  the  object  of  transformation?     What  is  Case  I?     Transposi 
tion?     Give  analyses. 

10 


110  SIMPLE   EQUATIO^^S. 

It  will  be  seen  that  in  each  of  these  examples,  the  term 
transposed  disappears  from  the  member  in  which  it  is  given, 
and  reappears  in  the  other  member  with  the  opposite  sign. 
Hence  the  following 

KuLE.  Dy^op  the  term  to  he  transjiosed  from  the  member 
in  which  it  stands,  and  write  it  in  the  other  member  with  its 
sign  changed, 

EXAMPLES   FOR   PRACTICE. 

In  the  following  equations  transpose  the  unknown  terms  to 
the  first  member,  and  the  known  terms  to  the  second. 

3.  dx  +  m  =  b.  •  Ans.  3x  =  b  —  m. 

4.  4:x  =  2x  -\-  cd.  Ans,  4:x  —  2x  =^  cd, 

5.  3m  4-  12.C  -—  c  =  ic  +  d 

Ans.  \2x  —  X  ^=  d  —  Sm  +  c, 

6.  —  bc'^d  +  ax  =  —  bx  —  m. 

Ans.  ax  +  bx  =  bc'^d  —  m. 

7.  4ac'x  —  od^  =  a^d  —  d^x, 

Ans.  4:acx  +  d^x  =  d^d  +  ScZ"^ 
8    a  +  b  —  x  —  c'  =  2c  —  bx. 

Ani!^  bx  —  x  =  2g  —  a  —  b  +  c\ 
9.  a^x  —  cd  =  b  +  a'^x  —  ax. 

Arts,  a^x  —  a^x  +  ax  ==b  -\-  cd. 

10.  ab  —  xc  =  bed  —  g.      Ans.  — xc==  bed  —  g  —  ab, 

11.  m  =  ax  —  Sex  +  m\       Ans.  Sex  —  ax  =  m^  —  m 

12.  0  =  ab—Scx  —  2ax-^c, 

Ans,  Sex  +  2aa;  =  ab  —  o. 

CASE  IT. 

IS©.  To  clear  an  equation  of  fractions. 

We  have  seen  (119,  II),  that  if  a  fraction  be  multiplied  by 
auj  multiple  of  its  denominator,  the  prodiu't  will  be  an  entire 


,1 


Give  Rule.     Case  II. 


ONE    UNKNOWN    QUANTITY.  m 

quantity.  Ilence  it  follows,  that  if  several  fractions  be  multi- 
plied by  a  common  multiple  of  their  denominators,  all  the  pro- 
ducts will  be  entire  quantities. 

1.  Transform  —  -f  —  =  3  into  an  equation  having  no  frac- 
tional terms. 

OPERATION,  Analysis.     We  multiply  every  ^erm  of 

^  the  equation  by  12,  the  least  common  mul- 

^.   --  =  3  tiple  of  the  denominators,  canceling    each 

^  ^  denominator  as  we  proceed ;  thus,  4  is  con- 

9x  -\-  2x  =  36  tained  in  12,  3  times,  and  3  times  Zx  is  9x ; 

6  is  contained  in  12  twice,  and  2  times  x  ia 

2x;  and  passing  to  the  .second  member,  12  times  3  is  86.    Hence,  the 

KuLE.  Multiply  all  the  terms  of  the  equation  by  the  least 
common  multiple  of  the  denominators,  i^educing  each  frac- 
tional term  to  an  entire  quantity  by  cancellation.  ♦ 

Notes.  1.  If  a  fraction  have  the  minus  sign  before  it,  change  all  the 
signs  of  the  numerator,  when  the  denominator  disappears. 

2.  The  pupil  will  readily  see  that  an  equation  may  also  be  cleared  of 
fractions  by  multiplying  each  term  by  all  the  denominators. 

EXAMPLES  FOR  PRACTICE. 

Clear  the  following  equations  of  their  fractions  :  — 

2.  ^  +  ^  =  11.  Ans,  3a:  +  Sx  =  132. 

3.  5 _ ^=  a,  /        Ans.  lOx  —  3a;  =  12<ju- 

o        4  /  ^ 

.    X       X       2x       ^ 

I  Ans.  5x  +  2x  —  2x  =  30a  —  10m. 

6    ^_^!  =  "  Ans.  3x  —  Qx  + IS  =  2a. 

'•-^ 
6.  |+|+J=^.  Ans.  6x  +  Bx  +  4x  =  10. 


Give  analysis.     Rule. 


.2 

SIMPLE  EQUA'l'IONS. 

7, 

|--+«       14-^t"- 

Ans.  2x  -f-  10a  —  14C  =  5a;  +  5a. 

8. 

X        « —  «^  _  1          ^         If.        rr4-'7-!n'--2 

9. 

aj  — 4       2aj5a;  — 3 

3       '     7        14           2     • 

Ans    Ux  —  56  -{-  12a:  =  15  —  21a;  +  68. 

10. 

a;         X  —  7         - 

.  0'  "  -4ns.  2a;  +  ax  —  7a  =  2Gri[*  —  lao. 

11. 7  -\'  c  •=^  fn,      Ans.  a?  -f-  ac  +  5c  =»  am  +  but 

a  -f-  6 

*   a  -f  ^        « —  2)  *. 

^ns.  ax  —  hx  —  ax  —  hx  =  4(a^  •««.  5^. 

13.  ^  +  :^  +  ;^  =  1.  -4^s.  ^/+  hf+  bd  =  5e?fa;. 
oa?       cix       raj 

^  .         a;  a;  1 

14.  — r-T  + 


Ans,  ax  —  bx  -^  ax  4.  5a;  =  1. 
-,  _    a.       a?       a;       a;       »»■» 
1^-  2  +  3  +  4+5  =  ^'- 

Jns.  SOa;  ^  20a;  +  15a;  +  12a;  ==  4620 

16.  j|  —  ''~^^  =  20.         Ans.  9a;  —  2a;  +  30  =  840. 

3x       5^       8x  +  12  __  1^ 

"^ '•  Y  ~  3   "^        7        ""  14' 

Ans.  63a;  —  70a;  +  48a;  +  72  =  3. 
,^    2x       a;  — 1       3 

18.  ^ —r  +  -2  =  1  — a;. 

da  *iao  a'' 


Ans.  4,abx  —  3ax  +  3a  +  186  =  6a*6(l  —  x) 

1  — ~"  —  —""  771 

19.  1 =  2.  Ans.  12  —  3a;  —  12m  =  4x, 

20.  ?^  —  2^V  =  ^  tV  ^^«.  9a;  —  103  =  54. 


ONE   UNKNOWN   QUANTITY.  H3 

REDUCTION   OF   SIMPLE   EQUATIONS. 

15 7.  The  Reduction  of  an  Equation  is  the  process  of  find- 
ing the  value  of  the  unknown  quantity. 

To  reduce  an  equation,  we  must  so  transform  it  that  the  in- 
known  quantity  shall  stand  alone,  and  constitute  one  member ; 
the  other  member  will  then  be  the  value  of  the  unknown 
quantity. 

158.  The  value  of  the  unknown  quantity  is  said  to  be  vej^ir 
fied,  when,  being  substituted  for  the  unkno\\Ti  quantity,  the  two 
members  of  the  equation  prove  to  be  equal.  When  this  occurs, 
the  equation  also  is  said  to  be  satisfied. 

ISO,  The  unknown  quantity  of  an  equation  may  be  united 
to  known  quantities  in  four  different  ways ;  by  addition,  by 
subtraction,  by  multiplication,  and  by  division  ;  and  further  by 
various  combinations  of  these  four  ways,  as  shown  by  the  fol- 
lowing equations,  both  numeral  and  literal :  — 

NUMERAL.  LITEltAL. 

1st.  By  addition,     .     .     x4-6  =  10  ic-fa=6, 

2d.    By  subtraction,     .    x  —  8  =  12  x  —  c  ==^  d. 

8d.    By  multiplication,         20x  =  80  ax  =  e, 

4th.  By  division,      .  .       — =16  --==  g  ^  a. 

5th.  a:  +  6-8  +  4  =  10  +  2  —  3,  a:  +  a  —  6  +  c=^ 
+  c,  &c.,  are  equations  in  which  the  unknown  is  connected 
with  known  quantities,  both  by  addition  and  subtraction. 

6th.  2x  +  ^=  21,     ax  +  —  =  c,  are  equations  in  which 

the  unknown  is  connected  with  known   quantities,  by  both 
multiplication  and  division. 


What  19  the  reduction  of  an  equation  ?  In  what  does  it  consist  ?  The 
term,  verified,  i?  used  to  denote  what  ?  The  term,  satisfied,  what  ?  When 
and  hoAv  is  an  equation  reduced  by  addition  ?  By  subtraction  ?  Multi- 
plication ?     Division  ? 

10^  H 


114 


SIMPLE  EQUATIONS. 


Equations  often  occur,  in  solving  problems,  in  wliich  all  of 
these  operations  are  combined. 

140.  When  the  unknown  quantity  is  united  to  known  quan- 
tities by  addition  or  subtraction,  it  may  be  disunited,  as  we 
have  seen,  by  transposition ;  which,  in  case  of  plus  or  posi- 
tine  terms,  is  equivalent  to  a  process  of  subtraction,  and  in 
case  of  minus  or  negative  terms,  to  a  process  of  addition. 

When  the  unknown  quantity  is  united  to  others  by  division, 
it  may  be  disunited  or  cleared  of  fractions,  by  multiplication  : 
and,  when  united  by  multiplication,  it  may  be  disunited  by 
division, 

1.  In  the  equation,  4a:  =  20,  what  is  the  value  of  a;? 


OPERATION. 

4x  =  20 
X  =■    5 

VERIFICATION. 

4  X  5  =  20 
20  =  20 


Analysis.  Dividing  both  members  of  the 
equation  by  4  (134,  IV),  we  obtain  a;  =  5. 
To  verify  this  value  of  x,  we  multiply  it  by  4, 
the  coefficient  of  a:  in  the  given  equation,  and 
obtain  20  =  20,  in  which  the  members  are 
equal ;  and  the  value  is  therefore  verified,  and 
the  equation  satisfied. 


X         X 

Given  x  -{-  -^ —  —  =  7,  to  find  the  value  of  x. 


OPERATION. 

X             X               ^ 

x+   ^— -=    7 

(1) 

Ix  +  2x  —  5x  =  70 

(2) 

7a:  =  70 

(3) 

X  =  10 

(4) 

VERIFICATION. 

a.4»-L«., 

(1) 

10  +   2  ^^  5  =  7 

(2) 

7  =7 

(3) 

Analysis.  We  clear  the 
given  equation  of  fractions, 
and  obtain  equation  (2) ;  uni- 
ting similar  terms  we  have 
(3) ;  and  dividing  both  mem- 
bers by  7  we  obtain  x  =  10. 

To  verify  this  value  of  a;,  we 
write  the  value,  10,  instead 
of  X,  in  the  given  equation, 
and  obtain  (i)  for  a  result ; 
performing  the  operations  in- 
dicated, we  have  (2) ;  and 
collecting  the  terms  in  tho 
first  member,  we  have  7  =  7; 
and  the  value  of  .t  is  verified, 
and  the  equation  is  eatisfied. 


ONE   UNKNOWN    QUANTITY.  115 

8.  Reduce  the  equation,  —  =  6  —  x. 

OPERATION.  Analysis.     We   first   clear 

the  equation  of  frac^ons,  and 

—  =z  b    —  X,  (1)  obtain  (2) ;    transposing  — ca*, 

^  we   have   (3) ;   factoring  with 

ax  =  be  —  cx,  (2)  reference  to  Xy  we   have    (4) ; 

ax  ••}-  cx  =  be.  (3)  and  lastly,  dividing  both  mera- 

(a  +  c)x  =  be.  (4)  hers   of  (4)  by  a  +  c,  the  co- 


bo 


efficient  of  x,  we  have  for  tlie 


x  = .        (5)  1        r         i>c 

a  -{-  c  value  of  a:, 


VERIFICATION. 


a  -{-  c 
To   verify   this   value,   wo 
substitute  it  for  x  in  the  given 
?  V.   __i£_  j_.  h  (I)  equation,  and  obtain  (i);  por- 

C        a  +  C  a  +  c  forming    the    operations    in- 

ab  ab  dicated,  we  have  (2),  in  which 

cTa^c  ^^  a  4-  c'     ^^'  *^^®  *^^  members  are  identi- 

cal, and   the  value   obtained 
for  X  is  verified,  and  the  equation  Is  satisfied. 

From  these  examples  we  deduce  the  following 

Rule.  To  reduce  an  equation  :  —  I.  Clear  the  equation 
offraciionSj  and  perform  all  the  operations  indicated. 

II.  Transpose  the  unknown  terms  to  the  first  member  of 
the  equation,  aiid  the  known  terms  to  the  second  member,  and 
reduce  each  member  to  its  simplest  expression,  factoring, 
when  necessary,  xoith  reference  to  the  unknown  quantity. 

III.  Divide  both  members  by  the  coefficient  of  the  unknown 
quantity,  and  the  second  member  loill  be  the  value  required. 

To  verify  the  result :  —  Substitute  the  value  found  for  the 
unknown  quantity  in  the  given  equation,  and  perform  the 
operations  indicated.  If  the  result  is  an  identical  equation, 
the  value  is  verified. 

The  three  steps  in  the  reduction  of  a  simple  equation,  con- 
Give  the  Rule  for  reducing  an  equation.     For  verifying  the  result 
The  steps  in  a  rod  action. 


116  SIMPLE  EQUATIONS. 

taining  but  one  unknown  quantity,  may  be  briefly  stated  as 
follows :  — 

1st.    Clear  of  fractions. 

2d.     Transpose  and  unite  terms. 

8d.    Divide. 

Notes.  1.  It  is  often  advantageous  to  transpose  and  reduce,  in  part, 
before  clearing  of  fractions. 

2.  In  the  last  step,  when  the  coefficient  of  the  unknown  quantity  is 
negative,  dividing  will  give  a  positive  result. 


EXAMPLES  rOR  PRACTICE. 

4.  Given  bx  +  22  —  2x=  81,  to  find  x.  Ans,  x  =  S. 

5.  Given  4a;  +  20  —  6  =  34,  to  find  x,  Ans.  a;  ==  5. 

6.  Given  8a;  +  12  +  7a;  =  102,  to  find  x,  Ans.  a;  =  9. 

7.  Given  10a;  —  6a;  +  14  =  62,  to  find  x.  Ans.  x  =  12. 

8.  Given  bx  — 10  =  3a;  +  12,  to  find  x.  Ans.  x  ==  11. 

9.  Given  8a;  —  20  =  — x  —  4,  to  find  a;.  Ans.x=4t 

10.  Given  4a;  +  45  =  7a;  —  80,  to  find  x.      Ans.  x  =  25. 

11.  Given  x  —  1  =  4a; —  91,  to  find  x.        Ans,  x  =  80. 

12.  Given  8(a;  +  1)  +.4(a;  +  2)  ==  6(a;  +  3),  to  find  x. 

Ans.  a;  =  7. 

Note.  — First  perform  the  multiplications  indicated,  and  then  reduca 

13.  Given  5(a;  +  1)  +  6(a;  -f  2)  =  6(a;  +  7),  to  find  x, 

Ans.  a;  s=  5. 

14.  Give    7(a;  +  3)  — 4(3x_16)  =  45,  tofinda;. 

Ans.  X  «  8. 

15.  Gi^en  ax  -\-  bx=  am  -f  6m,  to  find  x. 

OPERATION.  Analysis.      We   factor  botli 

ax  +  hx   ^am-\-    bm  (i)            members  of  the  given  equation 

(a  +  ft>  =  (a    4-  b)m  (2)            and   obtain   equation   (2);     and 

^  dividing  bya-r  o,  the  coefncient 


a;  =  w  (3) 


of  Xt  wo  obtain  a;=m,  the  anjswer. 


DNE  UNKNOWN  QUANilTY.  117 

16.  Given  ex  —  a?  =  5c  -^  6,  to  find  x,  Ans,  x  =  b. 


a — c 


17.  Given  ax+  dx^  a — c,  to  find  x,      Atis,  x  =  ,. 

18.  Given  ax  -f-  w  =  ccc  -f  n,  to  find  x.  n —  m 

a —  c 

19.  Given  ax  —  bx  =  c  +  dx  —  m,  to  find  x, 

c  —  m 


Ans.  x  = 


a  —  b  —  df 


20.  Given  ^  +  16  =  J  +  J  +  17,  to  find  x. 


OPERATION. 
o^  XX  Analysis.    We  first  drop 

^IQ  =  ^  ^  -^  V7   (i)  16  from  both  members  of  the 


SX  XX 


equation,     and     obtain     (2) 


=  —   -f  —  -f  1     (2)  Then,   clearing  of  fractions, 

transposing  and  reducing,  we 


6a;  =  4rr  +  cc  +  8      (3)  ^ave  a;  =  8,  the  Ans. 

x=S  (4) 


X  X 

21.  Given  ^  —  3  +  —  =5  —  3,  to  find  x.    Ans.  a;  ==  6. 

22.  Given  ^  —  ~  4-  2  =  3,  to  find  a;.  Ans,  x  =  12. 

23.  Given  -j  +  g"  —  fi"  ~  19'  *^  ^^^  ^        ^^^*  aj  =  2 

24.  Given  -o"  +  j  =  -^  +  To»  to  find  x.      Ans.  x  =  38. 

25.  Given  -  +  ^^  +  26  =  S&,  to  find  x. 

^  ^  ^        2ab-\'6a 

Ans.  -^r-; . 

2  +  a 

26.  Given  ^  +  2^  +  11  =  j  +  17,  to  find  x. 

Ans.  X  =a  10. 

27.  Givfen  ^a:  +  ^x  +  ^vnd  3&,  td  find  the  value  of  a. 


XX             'X           ^.^ 

,+   3  +  4=39 

a) 

6x  +  4x  +  3:r=39x  12 

(2) 

13x  =  39  X  12 

(3) 

ic  =  3   X  12 

(4) 

or,          aj  =  36 

(5) 

118  SIMPLE  EQUATIONS. 

Analysis.  "We  multiply 
both  members  by  12,  the  least 
common  multiple  of  the  de- 
nominators, indicating  the 
operation  in  the  second  mem- 
ber, and  obtain  (2) ;  reducinji;, 
vre  have  (3) ;  we  next  divide  by 
13,  observing,  in  the  second 
member,  that  a  quantity  may 

be  divided  by  dividing  any  one  of  its  factors,  and  obtain  (4) ;  wlienca 

a?  ==36. 

28.  GiTen  \x  -f  ^x  4-  ^x-^-  ^x=  *11,  to  find  x, 

Ans,  X  =  60. 

29.  In  }^  +  ^x  i-|x  =  130,  find  x.  Ans.  x  =  120. 

30.  Given  ^x  +  ^oj  +  y*^^  =  ^^y  ^^  ^^^^  ^• 

Ans,  X  =  120. 

31.  Given  iy  +  Jz/  +  42/  =  ^2,  to  find  y. 

Ans,  y  =  84. 

32.  Given  5a;  -f  ^x  +  ^x  =  34,  to  find  x. 

Ans,  X  =  6, 

_„    _        3x      x—1       ^         20x  +  13  ^    .    _ 

33.  Given ^r —  =  6a: -. ,  to  find  x. 

4  2  4       ' 

iFIRST   OPERATION. 

2>x      57  —  1       ^         20x  +  13 


4            2           -•^'    ■        4 

U/ 

3a;  — 2a:+  2  =  24a;  — 20a?- 

-13 

(2) 

-_3a;  =  — 15 

(3) 

a;=  5 

(4) 

Analysis.  AVe  multiply  by  4  to  clear  of  fractions.  The  first  term 
multiplied  by  4  is  3a: ;  the  product  of  the  second  term  is  2a;  —  2 ,  but 
since  the  fraction  has  the  minus  sign  before  it,  this  quantity  must  be 
subtracted,  v^hich  is  done  by  changing  the  signs  of  the  terms,  thus 
■ —  2a;  +  2.  The  product  of  the  first  term  in  the  second  member  is 
24a; ;  the  product  of  the  second  term,  or  fraction,  in  the  second 
member  is  20a;  +  13,  w^hich  being  subtracted,  as  indicated  by  the 
sign  before  the  fraction,  becomes  —  20a;  — 13.  Ke3ucing,  we  obtain 


:ne  unknown  quantity.  119 

Note. — Yv«ung  operators  are  liable  to  the  mistake  of  omitting  to  change 
the  signs  of  all  the  terms,  when  a  fraction  having  the  minus  sign  before 
it,  and  a  polynomial  numerator,  is  reduced  to  an  entire  quantity.  Thia 
error  may  be  avoided  by  the  method  which  appears  below. 

SECOND   OPERATION. 

dx  +  20x  +  l^=:24cx  +  2x  —  2         (S) 

Sx  = 15  (4) 

a;  =  5  (6) 

Analysis.  We  first  transpose  the  fractions  having  the  minus  sign, 
and  obtain  (2)  in  v^hich  the  fractions  are  positive ;  then  clearing  of 
fractions  and  reducing,  we  have  a:  ^=  5  as  before. 


34. 

x—S      9       x+4 ^    ^ ^ 
Given  X ^ —  =  o q — »  *o  find  x. 

Ans.  x  =  2. 

35. 

x  +  2      x  —  S   ,  ^              x  —  1  ,    ^   , 
Given  — q J [-2  =  X ^ — ,  to  find  x. 

36. 

Ans.  x=7, 

4x  —  2      3a7  — 5       .,   ,    ^  ^ 

Given  — :pi rr^ —  =  1,  to  find  x,    , 

11              13                               Ans,  a?  =  6. 

37. 

X       x  —  2           a;    ,  13  ^    ^  ^ 
Given  g- o —  =  —  2  "^  T'              ^' 

Ans.  X  =  10. 

Find  the  value  of  the  unknown  quantity  in  each  of  th« 
following  equations  : 

OQ    ^    ,  ^    ,  ^       ^  .  12a 

2  "*"  3       i"  ~  ^^'  ^  "^  "IF* 

a?       a;       a;       -  ahc 

39.  -  +  -  +  —  =  1.  Ans.  X  = 


40. 


a       b       c         '  '  ac  +  6c?  -f  a5* 

1+x      1+a  ^  _1 


1S.0  SIMPLE  EQUATIONS. 

41.  —^ g ^  =  _1.  Ans.  a?=«13. 

42.  1  —  5-^^  +  8  +  1  —  10  =  100^13. 

o  4  O 

Ans,  X  =  120. 
15  1 

43  — To  — Q — ro  =  ^Tl-  -4ns.  j;=:2. 
a;  +  o      3a;  +  9         '  ^ 

44  -  -j (-  ~  =  2,  -47W.  X  =  -Ka^  +  6^  +  c^. 

X  OL  X 


45.  (a  +  b)  (a  —  b)  (a;  .^  m)  ==  fc^m.     Ans,  x  = 

11^  —  80      8^  —  5       . 

46.  g -—-  —  O.  Ans.  37  =  10. 

An    1  +  I  +  m      12(m  +  6) 

47.  ^  t  =        9  '  •     ^ns.  a;  =  20. 

48.  ^-3 +  2(^-3)  =  ^-^  + a. 

OPERATION.  Analysis.      To    sim- 

Put      a;  —  3  =  2/ ;  (i)         ^^^^"^  *^®  equation,   we 

represent   x  —  3  by  y, 

Then  y  +  2^/  =  — ~  +  o      (2)         ^"^  ^^e  equivalent  equa- 

^        4  tion    is     (2),     reducing 

12a  "which,  we    obtain    y  = 

y==-BB  ^        12a     ^ 

"oF*    -But  y  represents 

Or,        X  —  3  =  -^^  (4)         x—3 ;  and  restoring  this 


35 


value  in  (i),  we   obtain 


«.       Q    r    -^^^  ^s         <^)»   whence,   by    trans- 

posing,  a;  =  3  +  -35-. 

49.  ^+  ^^^+1)  =  3(x+  5)_20.     Ans-x^l 

£0.  ^_c  +  ^  +  5^^IZf}  =  2i    ^ns.  a;=  a  + 1. 

.-     _.        9a; +  20      4a;_12      a;   ^    ^  ^ 
51.   Given  — gg—  =  ^__  +  -,  to  find  x. 


ONE  UNKNOAVN  QUANTITY. 


121 


OPERATION, 

Ax 


9a7  +  20  _  4^7  —  12       x 
~M  "5^—4  "^  4 

36(4^  —  12) 
bx  —  4 
20(5;r  — 4)==36(4^'  — 12) 
5(5a;  — 4)  =  9(4:t;~12) 
—  11a;  =—88 
x^S 


20  = 
-4 


52. 


6x  +  7  .  7a;  — 13 
Given  — J-, h 


(1) 

(2) 

(3) 
(4) 
(5) 
(6) 


Analysis.  We  first 
multiply  by  o6,  and 
dropping  9x  from  both 
members,  we  obtain 
(2).  Next,  clearing  of 
fractions,  we  obtain  (3) ; 
dividing  by  4,  we  ob- 
tain (4) ;  performing 
the  multiplication  indi- 
cated, and  reducing,  we 
obtain  (5)  and  (6). 


2ar  +  4 


9 


6^  +  3 


,  to  find  X. 


6j;+  7 
9 

3 


+ 


OPERATION. 

7^—13       2a? +  4 


6a;  +  3  ■ 
7a?— 13 
^  ^^Tl  ' 
21a?  — 39 


3 

2x  +  4 


=  5 


2a7+  1 
21x  —  39  =  10a;  +  5 
11a;  =  44 
a?  =  4 


(1^ 

(2) 

(3) 

(4) 

(5) 
(6) 


Analysis.  "We 
first  multiply  each 
term  by  3,  which 
is  done  by  dividing 
each  denominator  by 
3,  and  obtain  (2) ; 
multiplying  again 
by  3,  and  reducing, 
we  obtain  (3) ;  clear- 
ing effractions  and 
reducing,  we  obtain, 
finally,  (6). 


Note. — By  clearing  eqoiations  of  the  simplest  denominators  tirst,  aa 
iTi  the  examples  just  given,  we  sometimes  avoid  not  ofaly  laborious  multi- 
plications, but  the  involution  of  the  unknown  quantity  to  higher  powers. 


53.  Given 


7a? +  16 
21 


ri h  -Q»  to  find  X. 


21 


Ans, 


64.  Given 


65.  Given 


7x  +  2a 

X  +  a 

21 

4a;  — 11   ' 

2ar+l 

402  — 3a; 

29 


11 


5,  to  find  X. 
Ans,  X  = 
471  —  Qx 


12 


43a 
8a  — 21* 


,  to  find  X, 
Am,  x  =  72. 


J22                                     SIMPLE  EQUATIONS.  I 

^^    ^.        18a?  — 19   .11^  +  21      9^  +  15  ,»    ,  \ 

56.  Given  — ^-  +  -g^^  =  — ^,  to  find  ;..  . 

I                                                                Q             I       -I  Q  I 

57.  Given ^  +  5  +  x  =  — -q— ,  to  find  :r.  ' 

Ans.  x  =  %.  I 

58.  Given  ~^^ —. — ^  =  —-rjr  + »  ^^  ^^^  ^'  ] 

Ans,  X  =  —  •  ! 

PROPORTION.  \ 

i 

141.  It  is  often  convenient  to  express  the  relations  of  alge-  ■ 

braic  quantities  in  the  form  of  a  proportion,  and  from  the  \ 

proportion  derive  an  equation.  ■■. 

For  this  reason  we  present  here  so  much  of  the  theory  of  ] 

proportion  as  will  enable  the  pupil  to  make  use  of  this  prac-  j 

tical  advantage,  reserving  the  full  discussion  of  the  subject  for  j 

a  subsequent  chapter.  ] 

14S.  Ratio  is  the  quotient  of  one  number  divided  by  an*  i 

B  i 

other.     Thus,  the  ratio  of  -4  to  ^  is  -r.  \ 

A 

143.  Proportion  is  the  equality  of  ratios.  i 

Thus,  if  -^  =  r,  or  ^  =  rv4 ;  1 

and         —=:  Tf  or  D  =  rG'y  j 

G  J 

then  the  four  quantities.  Ay  B,  (7,  and  D,  are  proportional,  j 

and  their  proportionality  is  expressed  thus :  i 

A:  B  ::  G  :  D,  I 

in  which  A  and  Z)  are  called  the  extremes^  and  B  and  C  the  I 

means.  \ 

Define  Ratio.     Proportion.     The  extremes.     The  meana.  ^ 


ONE  UNKNOWN  QUANTITY.  123 

If  in  the  place  of  B  and  D  we  write  their  values,  rA  and 
rCj  this  proportion  will  be 

A  :  rA  : :  C  :  rC  ]  and  we  have 
A  X  rC  =  rACy  the  product  of  the  extremes ;  and 
G  X  rA  =  rACf  the  product  of  the  means.    Hence, 

The  product  of  the  extremes  is  equal  to  the  product  of  the 
means. 

1    Given  2  :  x  :  :  Q  :  bx  —  4,  to  find  x, 

OPERATION.  Analysis.     The  extremes  are  2  and 

2  :  X  :  :  Q  :  bx 4  ^^  —  4,  and  their  product  is  10a;  —  8  ; 

-I  A    8  =  6a:  (1)  ^^^  means  are  x  and  6,  and  their  pro- 

A  ci  ^  duct  is  6a:;  and  since  these  products 

are    equal,   we    have    equation    (i). 
^'  =  -^  (^)  Reducing,  we  have  x  =  2. 

Hence,  to  convert  a  proportion  into  an  equation,  we  have 
the  following 

Rule.     Place  the  product  of  the  extremes  equal  to  the  prch 
duct  of  the  means. 

EXAMPLES  FOR  PRACTICE. 

2.  Given  x  :  25  :  :  60  :  3,  to  find  x.  Ans,  x  =  500. 

3.  Given  a: :  x  +  6  :  :  2  :  6,  to  find  x.  Arts,  a:  =  3. 

4.  Given  x  +  2  :  a  :  :  6  :  c,  to  find  x,    ,  ah       ^ 

Ans,  X  = 2. 

G 

5.  Given  a:  +  6  :  38  —  x  :  :  9  :  2,  to  find  a;. 

Ans.  X  =  30. 

6.  Given  x  +  4  :  x  — 11 :  :  100  :  40,  to  find  x, 

Ans.  X  =  21, 
7    Given  x  +  a:  x  —  a::c:c?,  to  find  x. 

Ans.  x  =  -^^ ~. 

c  —  d 

Rule  fer  changing  a  proportion  to  an  equation. 


124  SIMPLE  EQUATIONS. 

8.  Given  x  :2x  —  a  :  :  a  :  h,  to  find  x.  ,  a/ 

Ans,  X  — 


2a —  ti 
9.  Given  a:b:  :2y  :  dfio  find  y,  Ans.  y  =  ^j, 

10.  Given  a*  —  ac  :  ax:  :1  :  (d  —  6),  to  find  x. 

Ans.  x^=  (d  —  h)  (a  —  c). 

11.  Given  cc  :  75  —  a; :  :  3  :  2,  to  find  x.       Ans,  x  ==  45 


PROBLEMS. 

144.  A  Problem  is  a  question  requiring  the  values  of  un- 

known  quantities  from  given  conditions. 

145.  The  Solution  of  a  Problem  is  the  process  of  finding 
the  values  of  the  unknown  quantities. 

We  have  seen  that  equations  serve  to  express  the  relations 
of  algebraic  quantities,  and  to  determine  the  values  of  the  un- 
known. In  Algebra,  the  solution  of  problems  is  generally 
effected  by  means  of  equations. 

1.  If  from  five  times  a  certain  number,  24  be  subtracted, 
the  remainder  will  be  equal  to  16  ;  required  the  number. 


SOLUTION.  Analysis.    We  denote  the  number  \ 

required  by  x.     By  the  condition  of  ' 

Let  X  =  the  number.       j,,^  problem,  5  times  x  minus  24  is  I 

.        -^  equal  to   16,  which,  expressed   alge-  i 

'  ^  '  braically,  gives  equation  (i).  : 

Dx  =  ^\J      (2)  Reducing  this  equation,  we  have  x  \ 

a:  =    8      (3)  =8,  the  number  required.  ; 

In  this  problem,  the  condition  from  which  the  equation  is  i 

formed  is  clearly  expressed,  and  furnishes  the  equation  directly,  \ 

2,  A  merchant  paid  $480  to  two  men,  A  and  B,  and  he  j 

paid  three  times  as  much  to  B  as  to  A.     How  many  dollars  \ 

did  he  pay  to  each  ?  \ 


Define  a  problem.     The  solution  of  a  problem. 


ONE  UNKNOWN  QUANTITY.  125 

SOLUTION.  Analysis.     We  let  x  :*opre- 

Let  X  =  the  sum  paid  to  A.  ^?"'  *^^  ^"/^  f'^  *^  ^ '  ^^^ 

_  ^  .  1        T^  Since  he  paid  6  times  as  much 

ox  =  the  sum  paid  to  B.  ^^  j^  ^^  ^^  ^  .3^  ^^i^  represent 

4x  =  the  sum  paid  to  both..       the  sum  paid  to  B ;  and  bj  ad- 
T     3"j.Qn  dition,  \x  is  the  sum  paid  to 

*  both.     But   hy   one   condition 

X  =  120,  A's  share.      (2)        ^f  the  problem,  480  dolhirg  ?s 
3x  =  360,  B's  share.       (3)         the  sum  paid  to  both.     There- 
fore 4x  is  equal  to  480,  which 
expressed  algebraically,  is   equation   (i).      Reducing,  we  find   the 
value  of  a:,  or  the  sum  paid  to  A,  to  be  $120;  and  3  times  this  sum, 
or  3  times  the  value  of   x,  the  sum  paid  to  B,  $300. 

In  this  problem  the  equation  is  drawn  from  an  implied  con- 
dition. In  the  algebraic  notation,  4x'  is  the  sum  paid  to  both  ; 
and  in  the  enunciation  of  the  problem,  480  dollars  is  the  sum 
paid  to  both  ;  these  two  expressions  of  the  same  quantity  are 
put  equal  to  each  other,  to  form  the  equation. 

From  the  examples  given,  we  derive  the  following 

GENERAL   RULE. 

I.  Eepresent  one  of  the  quantities  whose  value  is  to  he  de- 
termined hy  some  letter  or  symbolj  and  from  the  known  rela- 
tions find  an  algebraic  expression  for  each  of  the  other  quan- 
tities, if  any,  involved  in  the  question.  ^ 

II.  Form  an  equation  from  some  condition  expressed  or 
implied,  hy  indicating  the  operations  necessary  to  verify  the 
value  of  the  unknown  quantity, 

III.  Reduce  the  equation. 

The  three  parts  of  the  rule,  or  the  three  steps  in  the  solu- 
tion of  a  problem,  may  be  named  as  follows : 

1st.  The  notation; 

2d.    The  equation ; 

8d.    The  reduction  of  the  equation. 

Note. — By  the  first  two  steps,  the  problem  is  translated  from  common 
into  algebraic  language;  andv>this  is  called  the  statement  of  the  problem. 

Give  general  Kul%for  solution  of  problems.     Steps  of  the  process. 


126  SIMPLE  EQUATIONS. 


PROBLEMS. 


8.  A  father  is  3  times  as  old  as  his  son,  and  the  difference 
of  their  ages  is  24  years ;  what  is  the  age  of  each  ? 

Ans.  Son's  age,  12  ;  father's  age,  36. 

4.  A  gentleman  purchased  a  horse,  a  chaise  and  a  harness, 
for  $230.  The  chaise  cost  3  times  as  much  as  the  harness, 
and  the  horse  $20  more  than  the  chaise  ;  what  was  the  cost  of 
each?  r  Harness,    $30. 

Alls  J  Chaise,      $90. 
(Horse,     $110. 

5.  Two  men  bought  a  carriage  for  86  dollars;  one  paid  26 
dollars  more  than  five  times  as  much  as  the  other ;  what  did 
each  pay  ?  Ans.  One  paid  10,  the  other  76  dollars. 

6.  A  man  had  six  sons,  to  whom  he  gave  120  dollars, 
giving  to  each  one  4  dollars  more  than  to  his  next  younger 
brother ;  how  many  dollars  did  he  give  to  the  youngest  ? 

Ans.  $10. 

7.  Three  men  received  65  dollars,  the  second  receiving  5 
dollars  more  than  the  first,  and  the  third  10  dollars  more  than 
the  second  ;  what  sum  did  the  first  receive  ?  Ans,  $15. 

8.  A  man  paid  a  debt  of  29  dollars,  in  three  different  pay- 
ments ;  the  second  payment  was  3  dollars  more  than  at  first, 
and  the  third  payment  was  twice  as  much  as  the  second  ;  what 
was  the  amount  of  the  first  payment  ?  Ans.  $5. 

9.  The  greater  of  two  numbers  exceeds  the  less  by  14,  and 
o  times  the  greater  is  equal  to  10  times  the  less  ;  wliat  are  the 
numbers  ?  Ans.  6  and  20. 

10.  Moses  is  16  years  younger  than  his  brother  Joseph,  but 
3  times  the  age  of  Joseph  is  equal  to  5  times  that  of  Moses  ; 
what  are  their  ages  ?  Ans.  24  and  40. 

11.  On  a  certain  day,  a  merchant  paid  out  $2500  to  three 
men,  A ,  B,  and  C ;  he  paid  to  B  $500  less  than  the  sum  paid 
to  A,  and  to  C  $900  more  than  to  A  ;  required  the  sum  paid 
to  A.  Ans.  $700, 


ONE  UNKNOWN  QUANTITY.  121 

12.  There  are  three  nnnibers  which  together  make  72  ;  the 
second  is  twice  as  much  as  the  first,  and  the  third  is  as  much 
as  both  the  others  ;  what  are  the  numbers  ? 

A71S.  1st  is  12;  2d,  24;  3d,  36. 

13.  A  man  paid  $750  to  two  creditors,  A  and  B,  paying 
4  times  as  many  dollars  to  B  as  to  A  ;  how  much  did  he 
pay  to  each?  Ans.  To  A,  $150  ;  to  B,  $600. 

The  last  problem  would  have  been  essentially  the  same  iu 
character,  if  any  other  sum  had  been  paid  out,  and  if  the 
money  had  been  distributed  to  the  two  men  in  any  other  ratio. 
We  may  therefore  make  this  problem  general  by  stating  it  as 
follows : 

A  merchant  paid  a  dollars  to  two  men,  A  and  B,  paying 
n  times  as  many  dollars  to  B  as  to  A ;  how  much  did  he 
pay  to  each  ? 

SOLUTION.  Analysis.    Let 

T    ,  xi  •  1  i      i  ^  denote  the  sura 

Let    X  =  the  sum  paid  to  A  ;  ^^^^  ^^  ^ .  ^,^^^ 

nx  =  the  sum  paid  to  B.  ^x  ^in  denote  the 


X  +  nx=  a  a)        6""*    P^id   to  B, 

(l+n)x^a  (2)        ^"^   ''trJl^ 

^  ^  sum  paid  to  both, 

X  =  =^-~ — ,  the  sum  to  A  ;  (3)         which,  by  the  con- 

14"^^  dition  of  tiie  pro- 

na       .,              .     -r.  blem,     must      be 

nx==^-^^,themmtoB.  m        ^qual  to  «,  as  ex- 

pressed  in  (i). 
Reducing,  we  obtain  the  value  of  x,  or  A's  part,  in  (3);  and  multi- 
plying by  n  we  obtain  the  value  of  nx,  or  B's  part,  in  (4). 


VERLFICATION. 

na  a  -{-  na 


To  verify  these  values,  we 

,  add  them,  and  find  that  their 

14- ^1       1-fn         1  -\-  n  gm-jjj  jg  ^^  ^\^q  whole  amount 

__^  (1  +  n)a        paid  to  both,  as  stated  in  the 
~     1  -f.  itT        problem. 
=  a 


128  SIMPLE  EQUATIONS. 

14.  My  horse  and  saddle  are  worth  $100,  and  my  horse  is 
worth  7  times  as  much  as  my  saddle ;  what  is  the  value  of 
each  ?  Ans,  Saddle,  $12^ ;  horse,  $87^. 

15.  My  horse  and  saddle  are  worth  a  dollars,  and  my  horse 

is  worth  n  times  as  much  as  my  saddle  ;  what  is  the  value  of 

each  ?  ^        o.   -m         ^         1  ^^ 

Ans.  Saddle, ;  horse,  = — ■ — . 

1  +  n  '  1  -f  71 

16.  A  farmer  had  4  times  as  many  cows  as  horses,  and  5 
times  as  many  sheep  as  cows,  and  the  number  of  them  all  was 
100  ;  how  many  horses  had  he  ?  Ans,  4. 

17.  A  farmer  had  n  times  as  many  cows  as  horses,  and  m 

times  as  many  sheep  as  cows,  and  the  number  of  them  all  was 

a ;  how  many  horses  had  he  ?        .  a  , 

'  *^  Ans,  -^ ; horses. 

i  -f  ^  +  ^^^^ 

18.  A  school-girl  had  120  pins  and  needles,  and  she  had 
seven  times  as  many  pins  as  needles ;  how  many  had  she  of 
each  sort  ?  Ans,  15  needles,  and  105  pins. 

19.  A  teacher  said  that  her  school  consisted  of  64  scholars, 
and  that  there  were  3  times  as  many  in  Arithmetic  as  in 
Algebra,  and  4  times  as  many  in  Grammar  as  in  Arithmetic  ; 
how  many  were  there  in  each  study  ? 

Ans.  4  in  Algebra ;  12  in  Arithmetic  ;  and  48  in  Grammar. 

20.  There  was  a  school  consisting  of  a  scholars  ;  a  certain 
portion  of  them  studied  Algebra,  n  times  as  many  studied 
Arithmetic,  and  there  were  m  times  as  many  in  Grammar  as 
in  Arithmetic ;  how  many  were  in  Algebra  ? 

Ans,  J— , . 

1  +  n  +  rnn 

21.  A  person  was  $450  in  debt.  He  owed  to  A  a  certain 
'  gnm,  to  B  twice  as  much  as  to  A,  and  to  C  twice  as  much  as 

to  A  and  B  ;  how  much  did  he  owe  each  ? 

Ans.  To  A,  $50  ;  to  B,  $100  ;  to  C,  $300. 


ONE    UNKNOWN   QUANTITY 


129 


22.  A  person  said  that  he  was  owing  to  A  a  certain  sura, 
to  B  four  times  as  much,  to  C  eight  times  as  much,  and  to  D 
gix  times  as  much ;  and  that  $570  would  make  him  even  with 
the  world  ;  what  was  his  debt  to  A  ?  Ans.   $30. 

23.  A  person  said  that  ne  was  in  debt  to  four  individuals, 
A,  B,  C,  and  D,  to  the  amount  of  a  dollars  ;  and  that  he  was 
indebted  to  B,  n  times  as  many  dollars  as  to  A  ;  to  C,  m  timea 
as  many  dollars  as  to  A  ;  and  to  D,  p  times  as  many  dollars 
as  to  A  ;  what  was  his  debt  to  A  ? 


Ans. 


dollars. 


1  +  n  +  771  4-  p 

24.  If  $75  be  divided  between  two  men  in  the  proportion 
of  3  to  2,  what  will  be  the  respective  shares  ? 

Analysis.  "We  express  the 
greater  share  by  x,  and  the 
smaller  share  by  75  —  x. 
By  the  conditions  of  the  pro- 
blem, these  shares  are  in  the 
proportion  of  3  to  2,  as  ex- 
pressed in  the  proportion. 
Converting  this  proportion 
into  an  equation  and  reducing, 
we  find  the  respective  shares 
to  bo  $45  and  $30. 

Analysis.  Since  the  shares 
are  in  the  proportion  of  3  to 
2,  they  may  be  represented  by 


SOLUTION. 

Let 

X  : 

=  the  greater 

share  ; 

75- 

—  X 

=  the  smaller  share  ; 

JU 

:  75. 

—  a;;:3:2 

2x 

=  225  — 3x 

(1) 

ox 

=  225 

C-^) 

X  ==  45,  greater ;  (3) 
75  —  X  =  30,  smaller.  (4) 


Let 


SECOND    SOLUTION. 

;  =  the  greater  share  ; 
;  =  the  smaller  share  ; 


bx  =  75 

(1) 

^=15 

(2) 

Sx  ==  45, 

greater ; 

(3) 

2x  =  30, 

smaller. 

(4) 

Sx  and 


The  sura   of  the 


shares,  5a;,  must  be  equal  to 
$75,  the  amount  divided; 
whence,  by  reduction,  we  find 
the  shares  to  be  $45  and  $30. 
as  before. 

KoTK.  — V/hen  proportional  numbers  are  required,  it  is  generally  most 
convenient  to  represent  them  by  only  one  letter,  with  coefficients  of 
the  given  relation  or  proportion.  Thus,  numbers  in  proportion  of  8  to 
4,  may  be  expressed  by  Sx  and  4x,  and  the  proportion  of  a  to  6  may  be 
expressed  by  az  and  bx.  This  avoids  the  use  of  a  proportion  in  the  solu- 
tion of  the  problem. 


130  SIMPLE    EQUATIONS. 

25.  Divide  $150  into  two  parts,  so  that  tlie  smaller  may  be 
to  the  greater  as  7  to  8.  Ans,  $70,  and  $80. 

26.  Divide  $1235  between  A  and  B,  so  that  A^s  share  may 
be  to  B's  as  3  to  2.  Ans.  A's  share,  $741 ;  B's,  $494. 

27.  Divide  d  dollars  between  A  and  B,  so  that  A's  share 
may  be  to  B's  as  m  is  to  n. 

Ans.  A's  share,  — - —  ;  B's,  . 

m  -f-  n  m  -\-  n 

28.  Two  men  commenced  trade  together ;  the  first  put  in 
$40  more  than  the  second,  and  the  stock  of  the  first  was  to 
that  of  the  second  as  5  to  4  ;  what  was  the  stock  of  each  ? 

Ans.  $200,  and  $160. 

29.  A  man  was  hired  for  a  year,  for  $100  and  a  suit  of 
clothes ;  but  at  the  end  of  8  months  he  left,  and  received  his 
clothes  and  $60  in  money,  as  full  compensation  for  the  time  he 
had  w^orked ;  what  was  the  value  of  the  suit  of  clothes  ? 

Ans.  $20. 

30.  Three  men  trading  in  company  gained  $780,  which 
must  be  divided  in  proportion  to  their  stock.  A's  stock  was 
to  B's  as  2  to  3,  and  A's  to  C's  as  2  to  5.  What  part  of  the 
gain  must  each  receive  ? 

Ans.  A,  $156;  B,  $234;  C,  $390. 

31.  A  field  of  864  acres  is  to  be  divided  among  three* 
farmers.  A,  B,  and  C,  so  that  A's  part  shall  be  to  B's  as  5  to 
11,  and  C  may  receive  as  much  as  A  and  B  together ;  how 
much  must  each  receive  ? 

Ans,  A,  135  acres;  B,  297;  and  C,  432. 

32.  Three  men  trading  in  company,  put  in  money  in  the  fol- 
lowing proportion  ;  the  first,  3  dollars  as  often  as  the  second 
7,  ard  the  third  5.  They  gain  $960.  What  is  each  man's 
Bhare  of  the  gain  ?  Ans.  $192  ;  $148  ;  $320 


ONE  UNKNOWN  QUANTITY.  X3l 

83.  A  man  had  two  flocks  of  sheep,  each  containing  the 
same  number ;  from  one  he  sold  80,  and  from  the  other  20  ; 
then  the  number  remaining  in  the  former  was  to  the  number  in 
the  latter  as  2  to  3.  How  many  sheep  did  each  flock  origi- 
nally contain  ?  A'ns.  200. 

84.  A  gentleman  is  now  25  years  old,  and  his  youngest  bro- 
ther is  15.  How  many  years  must  elapse  before  their  ages 
will  be  in  the  proportion  of  5  to  4  ?  Ans.  25  years. 

85.  There  are  two  numbers  in  the  proportion  of  3  to  4 ; 
but  if  2i  be  added  to  each  of  them,  the  two  sums  will  be  in 
the  proportion  of  4  to  5.     What  are  the  numbers  ? 

Ans.  72  and  9G. 

3G.  A  man's  age  when  he  was  married  was  to  that  of  his 
wife  as  3  to  2  ;  and  when  they  had  lived  together  4  years,  his 
age  was  to  hers  as  7  to  5.  What  were  their  ages  w^hen  they 
were  married  ?  Ans.  His  age,  24  years  ;  hers,  16. 

87.  The  difference  between  two  numbers  is  12,  and  tho 
greater  is  to  the  less  as  11  to  7  ;  what  are  the  numbers  ? 

Ans.  21  and  33. 

38.  The  difference  between   two  numbers   is   a,  and  the 

greater  is  to  the  less  as  m  to  n ;  what  are  the  numbers  ? 

.  ma  ^       na 

Ans.  aud . 

m  —  n  m  —  n 

39.  The  sum  of  two  numbers  is  20,  and  their  sum  is  to  theii' 
difference  as  10  to  1 ;  what  are  the  numbers  ? 

Ans.  9  and  11. 

40.  The  sum  of  two  numbers  is  a,  and  their  sum  is  to  their 
difference  as  m  to  7i ;  what  are  the  numbers  ? 

Ans.  Greater,  -^^ — —  ;  less,  ^^ — t^ ~ 

2m       '         '2m 

41    A  certain  sum  ot  money  was  put  at  simple  interest,  and 


nO  —  xiVZQ  —  xiiSiU 

(A) 

Put  120  =  a  ;  then 

ej  — a::a  +  6  — ir::8:  14 

(B) 

8a  +  48  — •  Sjc  =  14a  —  14x 

(1) 

6x  =  6a  —  48 

(2) 

X  =  a  —  8 

(3) 

or  ^  =  112 

(i) 

132  SIMPLE  EQUATIONS. 

iu  8  months  it  amounted  to  $120  ;  had  the  interest  continued 
14  months,  the  amount  would  have  been  $126.  What  was  the 
principal  ? 

SOLUTION.  Analysis.   We  let 

Let  X  =  the  principal.  ^  represent  the  prin- 

120  _  :r  =  int.  for  8  mo.  ?^P^H  ^^^'  '"^^^"^'^ 

To^  .   ,    /,      1  ^  iiig    It    from     each 

126  — X  =  mt.  for  14  mo.  ^      ^       .       ^^^ 

amount,  we  have  120 

—  Xy  the  interc-Jt  for 
8  months,   and   126 

—  Xf  the  interest  for 
14  months.  But 
since  interest  is  pro- 
portional to  the  time, 
we  have  proportion 
(A).       For     brevity, 

we  write  a  for  120,  and  obtain  (b).  Converting  this  proportion  into 
an  equation,  and  reducing,  we  have  x  =  a  —  8,  or  120  —  8,  or  112. 

Note. — The  artifice,  employed  above,  of  representing  a  numeral  by  a 
letter,  and  restoring  the  value  in  the  final  result,  is  of  much  use,  and 
gives  true  delicacy  to  algebraic  operations.  The  pupil  should  be  en- 
couraged in  its  use. 

42.  A  sum  of  money  placed  at  simple  interest,  in  13  months 
amounted  to  $113,  and  in  20  months,  to  $120.  Required,  the 
sum  at  interest.  Ans.  $100. 

43.  A  certain  number  diminished  by  45,  is  to  the  same 
number  increased  by  45,  as  1  to  31.      What  is  the  number  ? 

Ans.  48. 

44.  The  number  12  is  |  of  what  number?' 

SOLUTION. 

r^et    X  =  the  number.  Analysis.     We    represent  the  re 

g  quired  number  by  x ;  and  by  the  con- 

■^  =  12  (1)  dition  of  the  question,  f  of  x  must  be 

equal  to  12,  which  gives  equation  (i). 

Sx  =  48  (2)  Reducing,  we  find  x  =  16,  the  answer 

a;  =  16  (3) 

45.  The  number  a  is  |  of  what  number  ?  Ans.  ~. 

46.  The  number  21  is  f  of  what  number  ?  Am.  4^. 


SOLUTION. 

Let    X     =  the  number. 

a     =  130 

X         ,     X 

6     +7    =^ 

a) 

7^  f  6^  =  42a 

(2) 

ISx  =  42a 

(3) 

X  =  3y^a 

(4) 

or,        a?  =  420 

(5) 

ONE    UNKNOWN   QUANTITY.  Igg 

47.  The  number  21  is  the  —  part  of  what  number  ? 

m 

48.  The  number  a  is  the  —  part  of  what  number  ? 

»  ,        an 

Ans.  — . 
m 

49.  If  you  add  together  ^  and  4  oi*  a  certain  number,  the 
mm  will  be  130  ;  what  is  the  number  ? 


Analysis.  Representing  the  re- 
quired number  by  a:,  and  the  numeral 
130  by  tty  we  have  by  the  condition  cf 
the  problem,  equation  (i).  Keducing, 
we  have  x  =  3  ,\a  (4) ,  and  restoring  the 
value  of  «,  we  have  a;  =  420,  the  num- 
ber sought. 


50.  A  farmer  wishes  to  mix  116  bushels  of  provender,  con- 
sisting of  rye,  barley,  and  oats,  so  that  the  mixture  may  con 
tain  f  as  much  barley  as  oats,  and  ^  as  much  rye  as  barley  ; 
how  much  of  each  kind  of  grain  must  there  be  in  the  mixture  ? 

Note. — Instead  of  employing  the  literal  quantity,  a,  Tve  may  avoid  the 
labor  of  multiplying  numbers  together,  by  indicating  the  operation. 

Analysis.     We  let  x 

represent  the  number  of 

bushels  of  oats.     Then, 

by  the  conditions,  |  of  x, 

5x  ,       1      , 

or  -=-,  must  be  the  bar- 

1  A  1     c^^         ^^ 

ley ;  and  J  of  y,  or  t^, 

must  be  the  rye.  Put 
ting  the  sum  of  ail  equal 
to  116,  we  have  equa- 
tion  (1).  Multiplying  by 
14,  indicating  the  ope- 
ration in  the  second 
Biember,  we  have  (2) ;  and  reducing  we  have,  finally,  x  =  56. 
12 


SOLUTION. 

Let    X    =  oats  ; 

dx       ,     , 
y  =  barley ; 

bx 

.  +  ^f+?i-m 

a) 

Ux  +  10x+  507=116  X  14 

(2) 

29a:  =  116  X  14 

(3) 

a?=4x  14 

(4) 

or,                     X  =  56 

(5) 

134:  SIMPLE   EQUATIONS. 

51  Divide  48  into  two  such  parts,  that  if  the  less  be  divided 
by  4,  and  the  greater  by  6,  the  sum  of  the  quotients  will  bo  9 

Ans,  12,  and  H6. 

52.  A  clerk  spends  |  of  his  salary  for  his  board,  and  |  of 
the  remainder  in  clothes,  and  yet  saves  $150  a  year.  What  is 
tiis  salary  ?  Ans.   $lo50. 

58,  An  estate  is  to  be  divided  among  4  children^  in  the  fol- 
lowing manner :  to  the  first,  $200  more  than  |  of  the  whole; 
to  the  second,  $340  more  than  -I  of  the  whole  ;  to  the  third. 
$800  more  than  ^  of  the  whole ;  and  to  the  fourth,  $400  more 
than  ^  of  the  whole.     What  is  the  value  of  the  estate  ? 

Ans.  $-l-800. 

54.  Of  a  detachment  of  soldiers,  |  are  on  actual  duty,  ^  of 
them  are  sick,  i  of  the  remainder  absent  on  leave,  and  the  rest, 
which  is  380,  have  deserted ;  v/ hat  was  the  number  of  men  in 
tiie  detachment  ?  Aiis.  2280  men. 

55.  A  man  has  a  lease  for  99  years  ;  being  asked  how 
much  of  it  had  already  expired,  he  answered  that  §  of  the  time 
past  was  equal  to  |  of  the  time  to  come.  Required  the  time 
past  and  the  time  to  come. 

Assume  a  =  99.     Ans.  Time  past,  54  years. 

56.  It  is  required  to  divide  the  number  204  into  two  such 
parts,  that  -r  of  the  less  being  taken  from  the  greater,  the  re- 
mainder will  be  equal  to  -|-  of  the  greater  subtracted  from  4 
times  the  less.  Ans.  The  numbers  are  154  and  50 

Put  a  "  204,  and  restore  value  in  the  result. 

( 7.  Tn  the  composition  of  a  quantity  of  gunpowder,  the 
{litre  Vvas  10  pounds  more  than  |  of  the  whole,  the  sulphur 
44  pounds  less  than  J  of  the  whole,  and  the  charcoal  2  pounds 
less  than  \  of  the  nitre.  What  was  the  amount  of  gun- 
powder ?       / 


=t 


ONE  UNKNOWN  QUANTITY. 


135 


SOLUTION. 

Let  X  —  the  Vs  hole ; 

2x 

~-  +  IC  =  the  nitre  ; 

o 

X 

-  —  4|  =  the  sulphur ; 
^-r  +  -^  —  2  =  the  charcoal. 

3  ^6^2T^T  ^  '^'       ~ 
4a;  +  a:+-y-  +  ---+21 


a) 


=  6j:.     (2) 


Analysis.  Represent- 
ing the  whole  composi- 
tion by  0-',  we  obtain  ex- 
pressions for  the  nitre, 
sulphur,  and  charcoal, 
according  to  the  condi- 
tions of  the  question ; 
and  the  sum  of  these 
three  ingredients  must 
be  equal  to  rr,  the  whole 
quantity  (i).  Multiply- 
ing by  6,  we  have  (2) ; 
reducing,  gives  ("0 ;  mul- 
tiplying by  7  we  have 
('!) ;  reducing  again,  gives 
(5) ;  dividing  by  3,  we 
have  (G) ;  and  reducing 
still  again,  we  have  a;zz69, 
the  answer. 


58.  Divide  $44  between  three  men,  A,  B,  and  C,  so  that 
the  share  of  A  may  be  |  that  of  B,  and  the  share  of  B  j  that 
of  C.  Ans,  A,  $9  ;  B,  $15  ;  C,  $20. 

59.  What  number  is  that,  to  v/hich,  if  we  add  its  |-,  ^,  and 
^,  the  sum  will  be  ^0  ?  Ans.  24. 

60.  What  number  is  that,  to  which,  if  we  add  its  -g-,  -J-,  and 


1%^V21       =.. 

(3) 

4x  +  60  +  21x7=7a:. 

60+21  x7=3x. 

20+  7x7=  X. 

Or,                      69=  X, 

(4) 
(5) 
(6) 
(7) 

\,  the  sum  will  be  a  ? 


Alls. 


Via 

25* 


61.  In  a  certain  orchard,  i  are  apple  trees,  \  peach  trees,  ^ 
plum  trees,  100  cherry  trees,  and  100  pear  trees.  How  many 
trees  in  the  orchard  ?  Ans.  2400. 

62.  A  farmer  has  his  sheep  in  five  different  fields,  viz.  :  \  in 
the  first  field,  |-  in  the  second,  ^  in  the  third,  ^V  in  the  fourtli, 
and  45  in  the  fifth.     How  many  sheep  in  all  the  fields  ? 

Ans.  120 


136  SIMPLE  EQUATIONS. 

63.  After. paying  out  ^  and  |  of  my  money,  1  had  remain- 
ing $66  ;  liow  many  dollars  had  I  at  first  ?  Ans,  120 

64.  After  paying  away  —  and  —  of  ray  money,  I  had  a  dol- 
lars left     How  many  dollars  had  I  at  first  ? 

.                amn 
Ans, . 


65.  If  from  ^  of  my  height  in  inches  12  be  subtracted,  ^  of 
the  remainder  will  be  2.     What  is  my  height. 

Ans.  5  feet  6  inches. 

66.  A  young  man,  who  had  just  come  into  possession  of  a 
fortune,  spent  f  of  it  the  first  year,  and  |  of  the  remainder  the 
next  year,  when  he  had  $1420  left.     What  was  his  fortune  . 

Ans,  $11360. 

67.  A  person  at  play  lost  ^  of  his  money,  and  then  won  3 
shillings ;  after  which  he  lost  ^  of  what  he  then  had  ;  and, 
on  counting,  found  that  he  had  12  shillings  remaining.  How 
much  had  he  at  first  ?  Ans,  20  shillings. 

68.  A  person  at  play  lost  a  fourth  of  his  money,  and  then 
won  3  shillings ;  after  which  he  lost  a  third  of  what  he  then 
had,  and  then  won  2  shillings ;  lastly,  he  lost  a  seventh  of 
what  ho  then  had,  and  then  found  that  he  had  but  12  shillings 
remaining.     How  much  had  he  at  first  ? 

Ans.  20  shillings. 

69.  A  shepherd  was  met  by  a  band  of  robbers,  who  plun- 
dered him  of  half  of  his  flock  and  half  a  sheep  over.  After- 
w^ard  a  second  party  met  him,  and  took  half  of  what  he  had 
left  and  half  a  sheep  over ;  and  soon  after  this  a  third  party 
met  him  and  treated  him  in  like  manner ;  and  then  he  had  5 
sheep  left.     How  many  sheep  had  he  at  first  ?        Ans.  47. 

70.  A  man  bought  a  horse  and  chaise  for  341(a)  dollars. 
If  f  of  the  price  of  the  horse  be  subtracted  from  twice  the 


ONE  UNKNOWN  QUANTITY.  Ig'J 

price  of  the  chaise,  the  remainder  will  be  the  same  as  if  ^  of 
the  price  of  the  chaise  be  subtracted  from  3  times  the  price 
of  the  horse.     Required  the  price  of  each. 

Arts.  Horse,  $152 ;  chaise,  $189. 

71.  If  A  can  build  a  certain  wall  in  10  days,  and  B  can  do 
the  same  in  14  days,  what  number  of  days  will  be  required  to 
build  the  wall,  if  they  both  work  together  ? 

Analysis.  Let  x 
reprev«ient  the  number 
of  days  both  are  to 
work.  Since  A  will 
build  j'g-  of  the  wall 
in  1  day,  he  will  build 

X 

Ypj  of  it  in  X  days  ;  and 

since  B  will  build  ^V 
in  1  day,  he  will  build 

y-r   in   aj   days.      But 

since  the  wall  is  to 
be  comjMed  by  both 
working  x  days,  the  sum  of  these  two  fractions  must  be  equal  to 
unity ;  hence  equation  (i),  which,  reduced,  gives  5|-,  the  number  of 
days  required. 

72.  If  A  can  do  a  piece  of  work  in  a  days,  and  B  can  do 

the  same  in  b  days,  how  long  will  it  take  them  if  they  both 

work  together  ?  ab 

Ans. days 

a  -{-  b 

73.  A  can  do  a  piece  of  work  in  12  days,  and  B  can  do  the 
same  in  24  days ;  how  many  days  will  be  required,  if  they 
both  work  together  ?  Ans.   8. 

74.  A  laborer.  A,  can  perform  a  piece  of  work  in  5  days, 

B  can  do  the  same  in  6  days,  and  C  in  8  da^ys ;  in  wliat  time 

can  the  three  together  perform  the  sam?  work  ? 

Ans.  2-X  days 
12  * 


SOLUTION. 

Let 

X  = 

:  the  number  of  days ; 

X 

:    A'^ 

5  part  of  the  work ; 

X 

B's 
1 

part  of  the  work. 

XX 

10  "^14== 

a) 

7x  +  5x  = 

70 

(^) 

12x  = 

70 

(3) 

X  = 

^1 

(4) 

138  SIMPLE  EQUATIONS. 

75.  A  laborer  engaged  to  serve  for  60  days  on  these  condi- 
tions :  for  every  day  he  worked  he  should  have  75  cents  and 
his  board,  and  for  every  day  he  was  idle  he  should  forfeit  25 
cents  for  damage  and  board.  At  the  end  of  the  time,  he 
received  $25.  How  many  days  did  he  work,  and  how  many 
days  was  he  idle? 

SOLUTION.  Analysis.     Representing 

,        ,  .,,  the  idle  days  by  a:,  60 — x 

Let       X  =  days  he  was  idle  ;  ^^^^  ^^  ^^^  ^^^^^.^^  ^^^^^ 

60  —  a?  r=:  days  he  worked.  n^d  he  worked  the  whole 

45  _—  a: 25  (^)  ^^   days,   his    wages    would 

have  amounted  to  $45.     But 


X 


=  20  (2) 


for  every  day  he  was   idle, 

75  cents,  but  25  cents  in 
addition,  making  $1  a  day.  In  x  idle  days,  therefore,  he  lost  x  dollars. 
Consequently,  the  amount  due  him  was  $45  —  2:  dollars.  But  by  the 
conditions  of  the  problem,  the  money  due  him  K^as  25  dollars ;  hence 
we  have  (i),  from  which  we  find  he  was  idle  20  days,  and  worked 
40  days. 

76.  A  person  engaged  to  work  a  days  on  these  conditions; 

for  each  day  he  worked  he  was  to  receive  b  cents,  and  for  each 

day  he  was  idle  he  was  to  forfeit  c  cents.     At  the  end  of  a 

days  he  received  d  cents ;  how  many   days  was  he  idle  ? 

.        ah  —  d 
Ans,  -J— — . 
0  +  c 

77.  A  boy  engaged  to  convey  30  glass  vessels  to  a  certain 
place,  on  condition  of  receiving  5  cents  for  every  one  he  de- 
livered safe,  and  forfeiting  12  cents  for  every  one  he  broke. 
On  settlement,  he  received  99  cents ;  how  many  did  he  break  ? 

Ans.  3. 

78.  A  boy  engaged  to  carry  n  glass  vessels  to  a  certain 
place  on  these  conditions  :  he  was  to  receive  a  cents  for  every 
one  he  delivered  safe,  and  to  forftfit  b  cents  for  every  one  he 
broke.     On  settlement  he  received  d  cents ;  how  many  did  he 

Ans,  The  number  represented  by  -^ — - 
^  "^    a  -^  b 


TWO   UKKNOWN   QUANTITIES.  139 

SIMPLE  EQUATIONS 
CONTAINING    TWO    UNKNOWN    QUANTITIES. 

14:&.  Independent  Equations  are  such  as  cannot  be  reduced 
to  the  same  form,  or  derived  one  from  the  other ;  as  x-\-  Sy 
:=  a,  and  4x  -[-  %  =  ^-  Independent  equations  refer  to  the 
same  problem,  and  express  diiFerent  conditions  of  the  problem. 

14:7  •  We  have  seen  that,  in  order  to  find  the  value  of  any 
unknown  quantity  in  an  equation,  we  separate  it  from  the  other 
quantities,  and  cause  it  to  stand  alone  as  one  member  of  the 
equation.  But  if  the  equation  contain  two  unknown  quanti- 
ties, the  value  of  neither  can  be  determined  by  this  process. 
To  show  the  reason  of  this,  let  us  consider  th^  following  equa- 
tion : 

Transposing  y,  we  have 

:r  rr:  20  —  ^, 

in  which  x  is  still  undetermined,  because  its  value  in  the  second 
member  of  the  equation  contains  the  unknown  quantity,  y. 
Again,  transposing  x  in  equation  ^^\  we  have 

y  =  20r-x, 

in  which  y  is  still  unknown,  because  its  value  contains  the  un- 
known quantity,  x.     Hence, 

Two  iinhiown  quantities  cannot  he  determined  from  a  sin- 
gle equation. 

The  equation  given  above  expresses  this  condition :  viz., 
the  sum  of  two  numbers  is  20 ;  and  since  there  are  many  pairs 
or  couplets  of  numbers  of  which  the  sum  is  20,  x  and  y  can 
have  no  particular  or  exclusive  values.       The  equation  is  satis- 


Define  independent  equations.  They  always  refer  to  Avhat  ?  Why 
cannot  the  va'ues  of  two  unknown  quantities  be  determined  from  on« 
equation  1 


140  SIMPLE  EQUATIONS. 

fied  if  we  make  a?  =  1  and  y  =  19i  or,  a;  ==  2  and  y  =^  18, 

etc  *  for 

*  '  1  +  19  =  20,  2  +  18  =  20,  etc.  ' 

But  if  we  combine  another  equation  with  this,  as  x  —  2/  —  4, 
wliich  expresses  a  different  condition  :  viz.,  that  the  difference 
of  the  two  numbers  is  4,  then  only  one  Talue  for  x  and  one 
value  for  y  will  satisfy  both  equations,  or  answer  both  condi- 
tions.    To  find  these  values  we  may  proceed  thus : 

x-\-y  =  20  a) 

x  —  y=    4:  (2) 


By  addition,  2x  =  24,  or  a:  =  12 

Subtracting  (2)  from  (i),  2^/  =  16,  or  ?/  =    8 

And  8  and  12  are  the  only  numbers  whose  sum  is  20  and 
difi'erence  4.     From  this  result  we  learn  that 

Two  uvknoivn  quantities  can  he  determined  from  two 
independent  equations. 

To  effect  the  reduction,  we  must  derive  from  the  two  a  new . 
equation  containing  but  one  unknown  quantity.     This  opera- 
tion is  called 

ELIMINATION. 

148.  Elimination  is  the  process  of  combining  two  or  more 
equations,  containing  two  or  more  unknown  quantities,  in  such 
a  manner  as  to  cause  one  or  more  of  the  unknown  quantities 
contained  in  them  to  disappear. 

There  are  three  principal  methods  of  elimination  : 
1st,  Bi/  substitution;  2d,  By  comparison;  od,  By  addition 
and  subtraction. 

CASE   L 

149.  Elimination  by  substitution. 

1.  Given  2x  +  by  =  31,  and  Sx  ■+  2y  =  19,  to  find  the 
values  of  x  and  y. 

How  many  equations  are  required  that  the  values  of  two  unknowr* 
quantities  may  be  determined?  V/hy?  Define  elimination.  Name  tho 
methods.     Give  Case  I. 


TWO  UNKNOWN  QUANTITIES.  ^4^ 

OPERATION.  Analysis.  We  transpose 

5y  in  equation  (A),  and  di- 


2a;  +  52/  =  31  (A) 


vide  by  2,  and  obtain  (i), 


QX  -]-  Zy  =  iv  (B)        which  expresses  the   alge- 

'  Qi        p,  braic  value  ofa:.  Tins  value 

X  = ^ (1)        of  X  we  substitute  for  x  in 

QQ        -  _  ^  (B) ;   thus,    instead   of  3a:, 

^  +  2v  =  19  (2)        "^G  write  its  value,  3  times 

2  31— 5y       93—157/       ^  .  , 

93  —  15i/  -f  %  =  38  (3)       — 2~»  ^^  — 2~  ^  '^^^''^ 

—  111/  =  —  55  (4)        with   the   other   terms   of 

^  __  5  /gN         equation  (B)written  in  their 

^1 9^  order,  gives  (2),  an  equatiou 

X  =s    "^ (6)         containing    only   one    un- 

^  known  quantity,  y ;  there- 

'  05  =  3  (7)        fore  x  has  been  eliminated. 

Reducing  in  the  usual 
manner,  we  have  y  =  5.  Since  y  is  5,  5y  is  25 ;  and  substituting 
^his  value  in  the  second  member  of  equation  (i),  we  have  (6),  which 
gives  the  value  of  x  in  known  terms.  Ileducing,  we  obtain  a;  =  3. 
Hence,  the  following 

Rule.  I.  Find  the  value  of  one  of  the  unknown  quan* 
titles  in  one  of  the  given  equations, 

II.  Substitute  this  value  for  the  same  unknown  quantity 
in  the  other  equation, 

EXAMPLES   FOR  PRACTICE. 

o    n-         <Sx  +  2y=2S\  ^    .    .  , 

2.  Given  <        ,    /       o-,  r  to  find  x  and  y. 
L  x  +  4:y=  21) 

Ans,  X  =  b]  y  =:  4:. 

S    Given  {^^  ~  ^^  ^  2o}  *^  ^^^  ^  ^""^  2/- 

Ans,  a;  =  3  ;  y  =  7, 


4.  Given  i  ^         />  r  to  find  x  and  y, 

tx —   y=    6)  ^ 

Ans,  X  =  7 1  y  =  1, 
Give  Analysis.    Rule. 


SIMPLE  EQUATIONS. 


6    Given  |  ""^  _^  ^^  =  13  I  *^  ^"^  ^  ^^^  ^' 

Ans.  2/  =  3;  a*  =  14. 

6.  Given  |  ^    __  _Ji  [  ^^  find  y  and  ^. 


-4ns.  y 


/ .  Given  i         ,    .         OT  r  to  find  x  and  2:. 
C   ^  +  5^  =  27  3 

Ans.  cc  =  7  ;  2:  =  4. 
8.  Given  ]  4f  "n^  —  "6  1  ^^  ^^^  2/  and  z. 


Ans.  2/  ~  14  ;  2  =  10. 


CASE  n. 


130.  Elimination  by  comparison. 

1.  Given  Sx  +  2y  =  16,  and  4x  -i-  Sy  =  23,  to  find  the 
values  of  x  and  y. 


OPERATION. 

Sx  +  2y==  16 
4a?  +  3^/  =  23 


16 

—  2v 

x  = 

3 

ir  = 

23 

_3y 

4 

16- 

■2y_ 

23 

-3y 

3 

4 

64- 

-8y  = 

69 

_9y 

y  = 

5 

dx 


10  =  16 

x=2 


(A) 

(B) 

a) 

(2) 

(3) 

(4) 
(5) 
(6) 

0) 


Analysis.  Transposing  2?/  in 
(A),  and  dividing  by  3,  we  obtain 
(1).  From  (B),  in  like  manner, 
we  obtain  (2).  We  have  thua 
found  two  algebraic  values  for  a:, 
which  must  be  equal  to  each 
other.  (Ax.  7.).  Equating  these 
values,  w^e  have  (3),  an  equation 
containing  but  one  unknown 
quantity,  y;  consequently,  x  has 
been  eliminated.  Reducing  this 
equation,  we  obtain  y=5.  Since 
y  is  5,  2y  is  10  ;  and  substitutirg 
this  value  in  (A),  we  have  (6), 
which  reduced,  gives  x  =  2. 


Give  Case  XL     Analysis. 


TWO    UNKNOWN   QUANTITIES.  |43 

Hence,  tlie  following 

Rule.  I.  Find  the  value  of  the  same  unknown  quantity 
in  each  of  the  given  equations, 

II.  Form  an  equation  by  placing  these  values  equal  to 
each  other. 

EXAMPLES  FOR  PRACTICE. 

2.  Given   {^l_2yZ    4_\  ^^  ^^^  ^  ^^^  2/- 

Ans.  X  =  S]  2/  =  4.. 

3.  Given   i      .    .         ^  r   to  find  y  and  x, 

Ans.  2/  =  5 ;  x  =  1. 

.     _         S2x  +  5^  =  29")    ^    „  ^  , 

4.  Given  i  o    ___    i  t  ^^  ^^^  ^  ^^^  ^' 

Ans.  X  =  7  'j  2  =  3. 

5.  Given   ]^  ^  ""    ^j,  [■  to  find  z  and  ?/. 

L2z —     y  =    15) 

Ans.  z  =  lo'j  2/  «=  11. 

.    _  n2x  +  Uy  =  26\   ^     .    ,  . 

6.  Given  -s    ^  "^         o  f   to  find  cc  and  y. 

(.  6x —     2/=2> 

Ans.  cc  =  1 ;  j^  =  1. 

^    ^.  (-4^  +  122/ =5)    ^    ^  ^ 

7.  Given  i        ,     o         i  r  to  find  x  and  y. 

i  x+    2y  =  1)  ^ 

Ans.  aj  =  i ;  y  =  |. 

(^+^«14) 
U^  6  f 

8.  Given    (  >  to  find  x  and  y. 

(8'"^6""       )        ^ns.  a;  =  16;  y  =  12. 

9.  Given  (         r.        ^  /to  find  a;  and  y. 

I      ,  oa;  —  '^y  ___^\ 

\  6  /      Ans.  cc  =  3  ;  y  ==  2. 

Give  Rule, 


12x  +  ISy  =  96 
12x+    62/ =  48 

(1) 

(2) 

12y  ==  48 

y=  4 

(3) 
(4) 

144  SIMPLE  EQUATIONS. 

CASE  III. 

151.  Elimination  by  addition  or  subtraction. 

1.  Given  6cc  +  9y  =  48,  and  4x  +  2y  =^  16,  to  find  the 
values  of  x  and  y, 

OPERATION.  .  Analysis.     We    multiply   equa. 

tion  (A)  by  2,  and  obtain  (i) ;  wo 
6jc  +     9i/  =  48       (A)  next  multiply  equation  (b)  by  3,  and 

4x  -\-     2v  =  16       (B)  obtain  (2).     We  have  thus  obtained 

two  equations  in  which  the  coefti- 
cients  of  x  are  equal.  Subtracting 
equation  (2)  from  equation  (i),  mem- 
ber from  member,  the  terms  con- 
taining X  cancel,  and  we  have  (3), 
an  equation  containing  but  one  un- 
~T     ~     ^     ZTTfi  known  quantity,  y.     Reducing,  we 

d     ~     t       ^°^  ^^""^    y   ==   4.      Substituting    this 

Ax  —    0      (6)  value  of  y  in  (b)  we  have  (5),  which 

a;  =    2      (7)  reduced  gives  x  =  2, 

2.  Given  7x  —  2y  =  31,  and  bx  +  2y  =:  29,  to  find  x 
and  y. 

OPERATION.  Analysis.     Since  the  coefficients 

7x  _  22/  =  31    (A)     <>f  y' !"  *«  ^'^T  ^ff"""''  '''^ 

9  9Q  numerically  equal,  and  have  oppo- 

Ox  +  22/  =  29       (B)  gj^g  signs,  their  sum  is  zero,  and  y 

I2x  =  60       (1)  ^^^^  ^®  eliminated  by  adding  equa- 

X  z=    d      (2)  ^^^'^^  ^'^'^  ^^^  ^^^'  ^^"^^^^  ^y  ^^^^" 

. ber;  this  gives  equation  (i),  from 

25  +  2t/  =  29       (3)  which  we  obtain  x  =  5.     Substi- 

2y  =    4:      (4)  tuting  this  value  of  x  in  (b),  we  ob- 

y  =z    2      (5)  *^i^  ^^)'  ^'^^^^  Sives  2/  =  2. 

From  these  examples  we  derive  the  following 

KuLE.     I.  Multiply  or  divide  the  equations  by  such  num 

bers  or  quantities  that  the  coefficients  of  the  quantity  to  he  eli 

minated  shall  be  made  equal 


Give  Case  III.     Analyses.    Rule. 


TWO  unkinov/n  quantities.  l45 

I.   1/  these  coefficients  have  unlike  signs,  add  the  prepared 
eqvaHons ;   if  like   signs,  subtract   one  equation  from  the 

other. 

Notes.  1.  If  the  given  equf^.tions  require  to  be  mnltiplied,  find  the 
le!i!*t  cornm  )n  multiple  of  the  coefficients  of  the  quantity  to  be  eliminatwl, 
find  <livi(Je  it  by  each  coefficient;  the  quotients  will  be  the  least  iFy^  ci- 
phers that  can  be  used.  -^ 

2  If  the  coefficients  are  prime  to  each  other,  multiply  each  equation 
hy  lite  coefficient  in  the  other  equation. 

3.   If  necessary,  clear  the  equations  of  fractions. 


EXAMPLES   FOR   PRACTICE. 

^    ^.  {2x  -{-  4y=28")    ,    ^    , 

8.  Given   K  ,         ,/       o-i  ^   to  find  x  and  v. 

Ans,  a:  =  4  ;  y  =  5. 

4.  Given   i  ^  -.  o  c*   to  nnd  x  and  v. 

Llx —   2/ =  13)  ^ 

Ans.  a?  =  2  ;  t/  =  1. 

5.  Given   i  ^         ^^^       ^.^  r   to  find  x  and  y, 

C6^  -f  12?/  =  72)  ^ 

^ns.  X  =  2]  y  =^b. 

6.  Given   i  ^  -r^  r  to  find  x  and  z. 

<  /x  —  z  =  <9) 

^ns.  X  =  12  ;  z  =  5. 
^    ^.         C2a;  +  22/  =  28)        ^  ^ 

7.  Given  i  -  "^ fi7  f  ^^  ^^^  ^  ^^^  2/- 

-ins.  a:  =  10  ;  2/  =  3. 

a.  Given  i  -.         _         .  r  to  find  x  and  y. 
C  iDx  —  5?/  =  4  J  ^ 

^ns.  a;  =  I  ;  i/  ==  }. 

(  3aT       22/  __  .  ) 

9.  Given  •^y"^"3        ^  ^  to  find  a?  and  2/. 

(.   x— 2/    =^3  j[ns.  ar  = 


Ans.  ar  =  7 ;  2/  =  3. 


3  3        "" 


i- 


3 
2a7— ^^r  ^  =  21^ 

Ans,  a?  =»  12  ;  y  =«  6. 


10.  Given  ^      '  3    ^  |^  to  find  x  and  y. 


13 


X46  SIMPLE  EQUATIONS. 

152.  Of  these  three  methods  of  elimination,  sometimes  one 
is  preferable  and  sometimes  another,  according  to  the  relation 
of  the  coefficients  and  the  positions  in  which  they  stand. 

No  one  should  be  prejudiced  against  either  method  ;  and  in 
practice  we  use  either  one,  or  modifications  of  either,  as  the 
case  may  require.  This  precept  may  be  applied  in  the  solu- 
tion  of  the  following 


GENERAL   EXAMPLES. 

1.  Given  ]         ,   ,         ^  ^  r  to  find  x  and  y. 

Arts,  y  =  2]  j?  =  8. 

2.  Given  \  ^         /      no  r  to  find  x  and  y. 

l^x  —  42/ =  38  J 

Arts,  a;  =  6 ;  y  =  4:, 

3.  Given  ]  ^     .   /?         o  i  i  to  find  x  and  y. 

Ans.  a?  =  8 ;  y  =  S. 

4.  Given  \       .      .  «  -.  o  r  to  find  x  and  y. 

(  —  4:x  +  2y=  —  12  J  ^ 

Ans,  a?  =  4 ;  y  =  2, 

^    /^.        j'6j?  +  5i/  =  128)        -  ^ 

5.  Given  i  „      ,    .  oo  r  to  find  x  and  y. 

1 3a?  +  41/=    88  »  ^ 

Ans,  a?  =  8  J  y  =  16. 

^-  ^'^^''^  {  6^  +  52  =  82  }  *°  ^°"*  ^  ^°^  ^- 

^ws.  a;  =  7 ;  z  =  8, 

^-  ^'''''^  {  S  i  2y  =  10  }  *°  ^°^  ^  "°"^  J/- 

J[ns.  a:  =  4;  t/  =  5. 

-4728.  a?  =  8  ;  i/  =  6. 

f  4t/  4-    z  =  102  f 
9.  Given  |       Za    __    .  g  j-  to  find  y  and  z. 

Ans.  2/ =  24;  z=f^6. 


.     TWO  UNKNOWN  QUANTITIES.  147 

f   ^^       I  Qlf   —        T   ^ 

10  Given  -J  Z      ,    ^.^^       ^..  ^  to  find  x  and  v- 

(  8x  4-  lUy  =  26  )  *^ 

JjiS.   j;  =  2  ;   2/  =  !• 

11  Given  f  ^^  +  ^'^  ==  ^'^  }  to  find  y  and  x, 

^?78.   ?/  =  12;  x=ll 

12.  Given  |  f '  t  t^  ""  !i  1  to  find  :r  and  y, 

u4?7s.  cc  =»  24  ;  2/  =  6 

13.  Given  {  ."^  "!"  ^-"^  ""  ^  to  find  x  and  v. 

Ars.  cc  =  6 ;  y  =  4. 
f  ir  -f  7?/  =  99  ) 

14.  Given  \]y_^{^^  5^  j  to  find  x  and  t/. 

^ns.  a:  =  7  ;  ?/  =  14, 

15.  Given  <       5  19  "^  V  to  find  x  and  ?/. 

t      X     —     y     ==lo] 

-<4/is.  a;  =  24  ;  y  =  14. 

C  X  +  V  =  2a  ) 

16.  Given  i  ..,    r  to  find  a;  and  v- 

{x  —  y=^2h)  -^ 

^ns.  a7  =  a-f-&;  y  =  (i  —  5. 

Ux  — 22/==8(f  — 2a  K     ^^ 

17.  Given  i         ,     '^  .  o     .   o  7  r  to  find  x  and  v. 

Arts.  x=^c  +  2d]  y  =  a  +  2g. 

18.  Given  \  "">,  r  to  find  x  and  y, 

Lex  —  my  =  26  j  ^ 

.                 a  +  6  a  —  b 

Ans.  x  =^ ;.  y  ■■ 


G         "  m 


19.  Given  ]    '         /  [-to  find  x  and  v. 

(  aas  —  by  =  c  J  ^ 

^  b'  +  d"  b—c 

Ans.  X  =  —J— — :- ;  y  =  -— — 

a{b  4-  c)     ^       b  -\-  c 

I \-  ny  =m  +  n 

20.  Given  (    ^         ,  )  to  find  a?  and  y. 


)  mx  ,  nV         ,   ,     o  I 


Ana,  X  =  mn  ;  y  «=  — . 
n 


148  SIMPLE  EQUATIONS. 

TROBLEMS 
PRODL'CINQ  EQUATIONS  CONTAINING  TWO  UNKNOWN  QUANTITIES 

1«>3.  Many  of  the  problems  hitherto  given  require  the  de- 
termination of  more  than  one  unknown  quantity ;  but  the 
quantities  are  so  related  to  each  other  that  they  can  be  ex- 
pressed algebraically  by  the  use  of  a  single  letter.  The  solu- 
tion is  often  rendered  more  simple  by  using  as  many  letters  as 
there  are  quantities  to  be  determined.  If  two  letters  are  em- 
l^loyed  to  represent  unknown  quantities,  the  conditions  of  the 
problem  must  furnish  two  independent  equations ;  otherwise 
it  will  not  be  capable  of  solution  (147). 

1.  A  man  bought  at  one  time  3  bushels  of  wheat  and  five 
bushels  of  rye  for  38  shillings ;  and  at  another  time,  6  bushels 
of  wheat  and  3  bushels  of  rye  for  48  shillings.  What  was  the 
price  of  a  bushel  of  each  ? 

SOLUTION.  Analysis.     We   represent 

.  the  prices  by  a;  and  y.    Since 

Let  j;==  price  of  wheat;         3   b^^hels  of  wheat  and  5 

y  =  price  of  rye.  bushels  of  rye  cost  38  shil- 

3^  +     by  =  38      (A)  li"g8,  we  have  equation  (A), 

Q^    ,      3^  =  48     (B)  furnished  by  the  first  condi'. 

-— — -  tioD  ;  and  since  6  bushels  of 

bx+Wy=  ib      a)  ^heat  and  3  bushels  of  rye 

(>x  -f     3i/  =  48      (2)  cost   48   shillings,    we   have 

•7^  ___  23      (3)  equation  (B),  from  the  second 

condition.       Multiplying   (A) 

by  (2)  to  make  the  coefficients 


2/  ==  4         (4) 


Bx  -|-  20  =  38      (5)  of  X  equal,  equations  (A)  and 

^  =  6         (6)  (B)  become  (l)  and  (2).     Sub- 

tracting (2)  from  (1),  we  have 
(3),  which,  reduced,  gives  y  =^  4.  Substituting  this  value  of  y  in  (A), 
W9  have  (5),  which  gives  t/  =  Q, 

2.  A  man  spent  30  cents  for  apples  and  pears,  bnying  his 
apples  at  the  rate  of  4  for  a  cent,  and  his  pears  at  the  rate  of 
5  for  a  cent.  He  afterward  let  a  friend  have  half  of  hia 
apphs  and  one-third  of  his  pears,  for  13  cents,  at  the  same 
rate.     How  many  did  he  buy  of  each  sort  ? 


TWO  UNKNOWN  QUANTITIES.  149 

SOLUTION. 

Let  X  =  number  of  apples,  and  y  =  number  of  pears. 
Hence,    j  —  cost  of  apples,  and  j  =  cost  of  pears. 

By  the  first  condition,      ►-  4.   |.  ,rr  30       ca) 

4:  O 

By  the  second  condition,  ^  +  :^  =  13        (B) 

o  10 


a) 


Multiplying  (b)  by  2,  |  +  j|  -_    26 

Subtracting  (i)  from  (a)  ^  —  ^  ^    4  ,2, 

Reducing,  3/  =  60  (3) 

Substituting  in  (A),  +    12  r:::i:  SO  (4) 

4 

Reducing,  x  _-=:  72  (s) 


8.  What  fraction  is  that,  to  the  numerator  of  which,  if  1  be 
lidded,  its  value  will  be  4,  but  if  I  be  added  to  the  denomina- 
tor, its  value  will  be  |  ? 

SOLUTION. 

Let  —  --  the  fraction. 

y 

By  the  first  condition,        -i—'—  ==  _      (a; 

y        *> 

X  1 

By  the  second  condition,  --.  =  -/      (B) 

?/  -f  1         4 

Clearing  (A)  of  fractions,    ?yx  +  3  =  t/  (i) 

Clearing  (i:  of  fractious,    ^c —  1  =:=   y  (2) 

Siil)tracting  (i;  from  (2;^      x  —  4  znz   0  (3) 

Trans[)()sing,  cc  r--   4  (4) 

From   (1)  y  r-  15  (S) 

X  ^ 

Hence,  tlie  fraction  is  -  r^  —     (6) 

y       lo 

18* 


150  SIMPLE  EQUATIONS. 

4.  .A  2:eiitlenian  paid  for  6  pairs  of  boots,  and  4  pairs  of 
shoes,  $41;  and  afterward,  for  o  pairs  of  boots,  and  7  pairs 
of  shoes,  $o2  ;  what  was  the  price  of  eaeh  per  pair  ? 

Ans.  Boots,  $6,  shoes,  fl. 

5.  What  fraction  is  that,  to  the  numerator  of  which,  if  4 
1)6  added,  the  value  is  ^;but  if  7  be  added  to  its  denominator, 
its  value  is  i  ?  Ant^.   j\, 

6.  A  and  B  have  certain  sums  of  money  :  says  A  to  B. 
'*Give  me  $15  of  your  money,  and  I  shall  have  five  times  as 
much  as  you  will  have  leff  Says  B  to  A,  "Give  me  $5  of 
your  money,  and  I  shall  have  exactly  as  much  as  you  will  have 
left ;"  how  many  dollars  has  each  ? 

Ans.   A  has  $35;  and  B,  $25. 

7.  What  fraction  is  that  whose  numerator  being  doubled, 
and  its  denominator  increased  by  7,  the  value  becomes  §  ;  but 
tlie  denominator  being  doubled,  and  the  numerator  increased 
by  1^,  the  vahie  becomes  |?  Ans.   |. 

8.  If  A  give  B  $5  of  his  money,  B  will  have  twn'ce  as  much 
money  as  A  has  left ;  and  if  B  give  A  $5,  A  will  have  thrice 
as  much  as  B  has  left ;  how  murh  had  each  ? 

Ans.  A  has  $13  ;  and  B,  $11. 

9.  A  merchant  has  sugar  at  9  cents  and  13  cents  a 
pound,  and  he  wishes  to  make  a  mixture  of  100  pounds  that 
shall  be  worth  12  cents  a  pound  ;  how  many  pounds  of  each 
quality  must  he  take? 

Ans,  25  pouTids  at  9  cents,  and  75  pounds  at  13  cents. 

10.  A  person  has  a  saddle  worth  £50,  and  two  horses. 
When  he  saddles  the  poorer  horse,  the  horse  and  saddle  are 
i^'orth  twice  as  much  as  the  better  horse  ;  but  when  he  saddN's 
the  better  horse,  the  horse  find  saddle  are  together  worth  thrt'o 
timee  the  other ;  what  is  the  value  of  each  horse  ? 

Ans.   £40  and  £KU. 

11.  One  day  a  gentleman  employs  4  men  and  ^  boys  to 
labor  for  him,  and  ])ays  them  40  shillings;  tlie  next  day  he 
hires  at  the  same  rate,  7  men  and  (i  boys,  for  50  shillings; 
what  are  the  daily  wages  of  each? 

Ans.  Man's,  5  shillings ;   boy's,  2  shillings  6  pence. 


TWO  UNKNOWN  QUANTITIES.  ][5| 

12.  A  merchant  sold  a  yard  of  broadcloth  and  3  yards  of 
velvet,  for  $25  ;  and,  at  another,  time,  4  yards  of  broadcloth 
and  5  yards  of  velvet,  for  $85  ;  what  was  the  price  of  each  per 
yard  ? 

Ans.  Broadcloth,  $10  ;  velvet,  $5. 

13.  Find  two  numbers,  such  that  half  the  first,  with  a  third 
part  of  the  second,  make  9  ;  and  a  fourth  part  of  the  first,  with 
a  fifth  part  of  the  second,  make  5. 

Ans.  8  and  15. 

14.  A  gentleman  being  asked  the  age  of  his  two  sons,  an- 
swered, ^'  If  to  the  sum  of  their  ages  18  be  added,  the  result 
will  be  double  the  age  of  the  elder ;  but  if  6  be  taken  from 
the  difference  of  their  ages,  the  remainder  will  be  equal  to  the 
age  of  the  younger;"  what  were  their  ages  ? 

Ans.  30  and  12. 

15.  A  says  to  B,  "  Give  me  100  of  your  dollars,  and  I  shall 
have  as  much  money  as  you."  B  replies,  "Give  me  100  of 
your  dollars  and  I  shall  have  twice  as  much  as  you  ;"  how 
many  dollars  has  each  ? 

Ans.  A,  $500 ;  B,  $7aO, 

16.  Find  two  numbers,  such  that  §  of  the  first  and  |  of  the 
second  added  together,  will  make  12  ;  and  if  the  first  be  divided 
by  2,  and  the  second  multiplied  by  3,  |  of  the  sum  of  these 
results  will  be  26 

Ans.  15  and  10|. 

17.  Says  A  to  B,  "^  of  the  difference  of  our  money  is 
equal  to  yours ;  and  if  you  give  me  $2,  I  shall  have  five  times 
as  much  as  you  ;"  how  much  has  each  l^ 

Ans,  A,  $48  ;  B,  $12. 

18.  A  market-woman  bought  eggs  to  the  amount  of  65 
cents,  some  at  the  rate  of  2  for  a  cent,  and  sonio  at  the  rate 
of  3  for  2  cents.  She  afterward  sold  them  all  for  120  cents, 
tlieroby  gaining  a  half  cent  on  each  egg ;  how  many  of  each 
kind  did  she  buv  ? 

Ans.  50  of  one :  60  of  the  other 


j^52  SIMPLE  EQUATIONS. 

19.  What  two  numbers  are  those,  whose  sum  is  a  and  dif- 


ference b  t 

SOLUTION. 


Let    X  =  the  greater. 
And  y  =  the  less. 


From  the  first  condition, 

X 

+  y=a 

a) 

From  the  second  condition, 

X 

—  y  :=b 

(2) 

Adding  (2)  to  (i), 

2x=^a  +  b 

(3) 

Subtracting  c^).  from  0), 

2y=za  —  b 

(4) 

Ilence, 

a    .    b 

greatei 

And 

a       b 

less. 

In  these  results,  we  have  the  algebraic  expression  of  the  fol- 
lowing general  truth : 

The  half  sum  of  any  two  numbers,  a<Med  to  the  half  dif- 
ference, is  the  greater  of  the  two  numbers;  and  the  half 
sum,  diminished  by  the  half  difference,  is  the  less 

20.  The  sum  of  two  numbers  is  28,  and  their  difference  is  6  ; 
what  are  the  numbers  ? 

Ans.  Greater,  17 ;  less,  11. 

21.  There  are  two  numbers  whose  sum  is  100,  and  three 
times  the  less  taken  from  twice  the  greater,  gives  150  for  re- 
mainder.    What  are  the  numbers  ? 

Ans.  90  and  10. 

22.  A  man  and  his  wife  labored  m  days,  and  received  2a 
dollars  for  compensation.  Had  the  wife  been  idle,  and  on  ex- 
pense at  the  same  daily  rate  as  her  wages,  they  would  have 
Baved  but  2c  dollars.     What  were  the  daily  wages  of  each  ? 

Ans.  Man^s, ;  wife's,  . 

Having  the  sum  aud  the  difference  of  two  numbers  given,  hew  uiuy  the 
D ambers  be  found? 


TKIIEE  OR  MORE  UNKNOWN  QUANTITIES.  I53 

23.  Find  two  numbers,  such  that  twice  tlie  first  climiuished 
by  the  second  shall  be  equal  to  ob ;  and  twice  the  second 
diminished  by  the  nrst  shall  be  equal  to  3^/. 

Ans.   First,  a  -f  2b )  second,  2a  +  b, 

'1  \ .  Divide  the  number  a  into  two  such  parts  that  the  firsV 

Hball  be  to  the  second  as  m  to  n. 

.        _.    ^       ma  .        na 

Ans.  First, ;  second, 

771  -h  n '  wi  4-  n 

SIMPLE  EQUATIONS 
CONTAINING    THREE    OR    MORE    UNKNOWN   QUANTITIES. 

fx  -f-    y  -^    z=    0^ 
X  +  2y  -\-  oz  =  IQ  >  to  find  x,  y,  and  z, 
^  +  3y4-4z  =  2lJ 

OPERATION.  Analysis.  By  tran* 

posino;,      we      obtain 
X  -jr     y  +     z  =    .)  (A)       equation  (i)   from  (A), 

X  +  2y  -i-  Sz  =  lQ  (B)        (2)  from  (B),  and  (3)  from 

X  -\-  ^y  -{■  4z  =21  (C)       (C).     These   equations 

~  give    us   three   values 

^=  ^ —  y —  2:  (1)  ^-^j.  ^^  Equating  the 
oc==l6  —  2y  —  Sz  (2)  1st  and  2d  values,  we 
a:  =  21  —  8?/  —  4z       (G)       have  (4),  and  equating 

the  2d  and  3d  values, 
we  have  (5).  We  have 
thus  eliminated  x,  and 
obtained  two  equations 
with  two  unknown 
quantities.  By  trans- 
^  ^^  '^  ^^^        posing  terms  in  (4)  and 

(9)       (5)   and    reducing,  we 
(10)       have  (6)  and  (7),  giving 
^j.        two  values  for//.  Equa- 
ting these  values,  we 
eliminate  y,  and  oh^iin  (8),  from  which  we  fiiwl  z  =  2.     By  substi- 


9 —    y —    z  =  16  —  2y  —  Sz 

(4) 

U^2y  —  Sz  =  21 -^Sy  —  4:z 

(5) 

y=   7  —  2z 

(6) 

y==    ^—   z 

(7) 

154  SIMPLE    EQUATIONS. 

tilting  the  value  of  z  in  (7)  we  obtain  ?/ r=r  3  ;  and  substituting  the 
values  of  z  and  y  in  (1)  w^e  obtain  x  ~  4. 

In  this  example  we  have  eliminated  bj  the  method  of 
comimri^on. 

C2x  +  4:y  —  nz  =  22^ 
2.  Given  <  4x  —  2i/  +  52  =  18  >  to  find  x,  y  and  z. 
\Qx  +  ly  —    2  =  63  J 

Analysts.     We  multiplj  ^a; 
by  2,  and  obtain  (i).    Writing 

(B)  underneath  this  and  sub- 
tracting, we  eliminate  x,  and 
obtain  (2).  Next,  multiplying 
(A)  by  3,  we  obtain  (3).  W^riting 

(C)  under  this  and  subtracting, 
we  again  eliminate  a:,  and  ob- 
tain (4).  Multiplying  (4)  by  2 
gives  (5).  Writing  (2)  under 
this  and  subtracting  (5)  from 
it,  we  eliminate  y  and  obtain 
(6),  which  reduced,  gives  2  =  4. 
Substituting  this  value  of  z  in 
(4),  and  reducing,  we  have 
y  =:  7  ;    and    substituting    the 

.  values  of  z  and  y  in  (A),  and 
reducing,  we  find  x  =  3. 

lu  this  example  we  have  eliminated  principally  by  the  method 
of  addition  or  subtraction. 

From  the  illustrations  given  we  deduce  the  follovfing 

Rule.  I.  Combine  one  of  the  equations  with  each  of  the 
others,  eliminating  successively  the  same  unknown  quantity  : 
the  result  will  he  a  new  set  of  equations  containing  one  less 
unknown  quantity. 

II.  Combine  one  of  the  new  equations  with  each  of  the 
others,  eliminating  a  second  unknown  quantity ;  and  the  re- 

Give  rule  for  reducing  equatians  containing  three  or  more  unknown 
quautitiei. 


OPERATION. 
2x  +42/—     32  =  22 

4x  — 2?/+    52=18 
^x  +  ly  —     2  =  63 

(A) 
(C) 

4.r  +  82/  —    62  ==  44 
4x  — 2^/4-    52=18 

(1) 

10?/— 112  =  26 

(2) 

6x+  12//—    92  =  66 
%x+    ly—     2=63 

(3) 

by  —    82  =    3 

(4) 

10?/— 162=    6 
10^  —  112  =  26 

(5) 

52  =  20 
2=    4 

y=  7 

x=  3 

(6) 
0) 
(8) 
(9) 

THREE  OR  MORE  UNKXOWN  QUANTITIES.       ^55 

suit  will  be  a  new  set  containing  two  unknown  quantities  lesn 
■Jian  the  oj-iginal. 

III.  Continue  this  iwocess  till  an  equation  is  found  con- 
taining but  one  unknown  quantity. 

lY.  Eeduce  this  equation  and  find  the  'jalue  of  the  un  - 
known  quantity.  Substitute  this  value  in  an  equation  con- 
taininj  two  unknown  quantities,  and  thus  find  the  value  of  a, 
second.  Substitute  these  values  in  an  equation  containing 
three  unknown  quantities,  and  find  the  value  of  a  third  ;  and 
80  on,  till  the  values  of  all  are  found. 

Note  — Instead  of  combining  the  first  witli  each  of  the  others,  we 
may  combine  the  first  with  the  second,  the  second  with  the  third,  and 
so  on ;  or  we  may  pursue  any  other  order  of  combination  best  suited  to 
the  mutual  rehitions  of  the  coefficients. 

loo.  It  is  evident  that  if  there  are  more  unknown  quanti- 
ties than  equations,  the  last  resulting  equation  will  contain 
tv7o  or  more  unknown  quantities,  and  the  solution  will  be  im- 
possible (147).     nence  the  following  general  law: 

There  must  be  as  many  independent  equations  as  there  are 
unknown  quantities. 

Note. — If  there  are  more  independent  equations  than  unknown  quan- 
tities, some  of  them  can  be  dispensed  with  in  reducing  the  equations. 

EXAIvrPLES   FOR   PRACTICE. 

1     Given  <     hx  +  \y  ~~  22  =  20  \  to  find  x,  y,  and  z. 
1 11^  +  7^  —  62  =  37  J 

Ans.  .T  ==  2  ;   y  =  3  ;   2  =  1 

r3j;-f  57/  +     ^=  -C)] 
2.  Given  <  G^  4.  S?/  -f  22  =  31  V  to  find  x,  y,  and  z. 
i.9^  4_  4^^^42=,  50  J 

Ans.  ^=2;  2/=3;  z=5. 
What  number  of  equations  is  required  in  the  solution  of  a  problem  ? 

Wuy? 


156  SIMPLE   EQUATIONS. 

Note  — AVhr^n  several  coefficients  are  unity,  or  multiples  of  enoh  otlieT, 
cert.iij)  exppdients  may  be  employed  to  facilitate  the  calculation,  for  wliicb 
uo  specitic  rules  can  be  given. 

(x  +  y  +  z^n^ 
3.  Given  <  x  -^  y  —  2  =  25  \  to  find  Xj  y,  and  z. 
lx  —  y  —  z=   9  J 

Subtract  the  2d  from  the  1st,  and  '2z  =  6,  or  2:  =3. 
Subtract  the  8d  from  the  2d,  and  2y  =  16,  or  3/  =  8. 
A«M  the  1st  and  3d,  and  2x  =  40,  or  x  =  20. 

ill  -i-  V  -\-  X  -{■  y  =  10  \ 
U-\-V-i-Z+X=ll)  ft,,,  ,  « 

{  to  find  the  value  of 

]      ^      ^      ^^       ^-(     each. 
Ju  +  x  +  y-{-z  =  lS\ 

\v-\-x-\-y  +  z  =  l-l] 

Since  in  each  equation  one  letter  is  wanting,  assume 
u  +  V  +  X  +  y  -{-  z  =  s 
Then  8  —  z  =  10 

s  —  y:=ll 

8  —  x  =  12       ^^^^®'      ^y 
8  —  v^lS 


substitu- 


By  addition,  56*  —  «  =  60 

6-=  15 


T  I  tins:     the 

value  of  Sy 


z  =  b 
y  =  i 
x  =  S 
v  =  2 
u=\ 


Required,  the  values  of  the  unknown  quantities  in  the  fol 
lowing  equations : 

rx4-?/  +  2=26^  Cx  =  12^ 

5.  Jx— -2/       —"^r  ^^^'  \y=  ^^ 

(x  —  z  =6j  (2=6, 

f   x  —  y  —  z=^    6^  Cx  =  ^^, 

6    <^  —  x  —  z=12\  Ans.    ly=r.2\^ 

^         llz-^y—x  =  2A)  U^12. 


'x  +  iy-lOO^  f^--64, 

7.    ^y+izr^rlOoV  Arts.    ly  =  12, 

^z  -1-^x1=:  100  J  (2  =  81. 


{i 


THREE  OR  MORE  UNKNOWN  QUANTITIES.  ^57 


=  52  \ 
=  82) 

'  =  68  \ 
=  30  ( 
=  32) 


z  -I-  1^  =  68  \  A718. 

If;  -f-  W  =  30  ' 
U  '\-  X 


^x  +  81/  =  23 

2/  4-  3^  =  81 
X  -f  2/  +  2  -f  2ifr  r=  39 

r  4x  4-  2y  —  82  =     4  ^ 

10.  .^  3x  —  5?/  4-  22  =  22  V  Arts. 

f  i^  +  ^!/  +  i^  -  22  ] 

11.  -{  i^c  -f    2/  +  i^  =  83  V  J72s. 

^   ^  +  ^2/  -  A^  ==  19  j 

12.  ^  .?:  4-  2  =  6  V  Ans,    -(?/==  ^(a  4-  c  —  6), 
( 2/  4-  z  =  c  J  (2=^64-0  —  a). 

I  .  .  \  /  « 

CJ^  -t    y  -{-  _az  =  a  -\-  ac  +  c      j  I   a;  =  -  , 

13.  /  ^'-^  +   i/  +  a'^2:  =  3«c  (   j^^g  )  y  =  ac, 

'  (  )  c 

ac^c  4-  2t/  +  ac2  =  o,'  4-  2ac  4-  cM  j  z  ==  — 

I  {  a 


PROBLEMS 

PRODUCING     EQUATIONS     CONTAINING     THREE     <R     M("RE 
UN  KNOWN     QUANTITIES. 

foO.  1.  Three  persons,  A,  B,  and  C,  talking  of  tlieir 
aires,  it  was  discovered  tliat  the  sum  of  their  ages  was  90  ;  the 
half  sura  of  A's  and  B's  was  25  ;  and  the  half  sum  of  B\s  and 
C's  was  35.     What  was  the  age  of  each  ? 

Ans.  A's  =  20  years ;  B's  =  80 ;  aud  C^s  =  40 
14 


158  SIMPLE  EQUATIONS. 

2.  A's  raoTiey  added  to  8  times  B's  and  C's  will  amount  to 
$170  ;  B's  money  added  to  4  times  A's  and  C^s  will  amount  to 
$580  ;  and  C's  money  added  to  5  times  A's  and  B's,  will 
amount  to  $ooO.     How  much  money  has  each  't 

SOLUTION. 


X  +  ?>y  +  ^z  = 

470. 

(1) 

By  the  conditions,          < 

y  +  4:x  +  4tzz= 

580. 

(2) 

1 

z  +  5x  +  5y  = 

680. 

(3> 

Adding  2.x  to  (i), 

'Sx  +  Sy  +  Sz=: 

470  -    2x 

(4) 

-            87/    to    02), 

4x  +  Ay  -\-  4:z  — 

580  +  '^y 

(5) 

''           42    to    (3),                 ( 

^  5x  +  by  4-  bz  = 

680  +  4z 

(6; 

Assume 

8  = 

X  +  y  +  z 

Equation  (4)  becomes 

r  Ss  = 

470  +  2x 

(7) 

(5) 

J    4s  = 

580  +  Sy 

(8) 

(G) 

(5.=. 

680  +  4z 

(9) 

Multiplying  (t)  by  6, 

^1Ss  = 

2820  +  12x 

(10)- 

''            (8)  by  4, 

.  Us  = 

2820  +  I2y 

(11) 

"           (9)  by  8, 

[Vos  = 

1890  +  122 

(12) 

Adding  (lo),  (U),  and  (i2), 

49s  = 

7080  4-  12s 

(13) 

S78  = 

7080 

(14) 

8  = 

190 

(15) 

Substituting  value  of  s  in  (T)  ,  Cx  =    $50,     A^s, 

"     "  («),  <y=      60,     B's, 

"  "         "     "  i^^ ,  Iz  =      80,     C's. 

Note. — The  pupil,  if  he  cboose,  may  solve  the  above  question  by  the 
u&unl  methods.  But  the  solution  given,  to  which  we  call  particular 
attention,  is  calculated  to  impart  a  superior  skill,  and  to  cultivate  a 
higher  mathematical  taste.  Similar  expedients  may  be  employed  in 
several  of  the  following  problems. 

8.  A  farmer  has  sheep  in  three  pastures.  The  niimlier  in 
the  fiist  pastuie,  added  to  half  the  number  in  the  second  and 
third,  will  make  70.    The  number  in  the  second  pasture,  added 


TJfllEE    OR   MCRE    UNKNOWN   QUANTITIES.  ^59 

la  one  third  of  the  number  in  the  first  and  third,  will  make  60. 
And  the  number  in  the  third  pasture,  added  to  one  fifth  of  the 
number  in  tlie  other  two,  will  make  58.  How  many  sheep  in 
each  pasture  ?     Ans.  In  the  first,  30  ;  second,  85  ;  third,  45. 

4.  Three  persons  divided  a  sum  of  money  among  them  in 
Buch  a  manner  that  the  shares  of  A  and  B  together  amounted 
to  $900,  the  shares  of  A  and  C  togetJier  to  $800,  and  the 
shares  of  B  and  C  to  $700  ;  what  was  the  share  of  each  ? 

Ans,  A's  share,  $500  ;  B's,  $400  ;  and  C's,  $300. 

5.  The  sum  of  three  numbers  is  59  ;  one  half  the  difi'erence 
of  the  first  and  second  is  5,  and  one  half  the  ditference  of  the 
first  and  third  is  9  ;  required  the  numbers. 

Ans.  29,  19,  and  11. 

6.  A  certain  number,  consisting  of  two  places,  a  unit  and  a 
ten,  is  four  times  the  sum  of  its  digits,  and  if  27  be  added  to 
it,  the  digits  will  be  inverted.     What  is  the  number  ? 

Note. — The  local  value  of  a  figure  is  increased  tenfold  by  every  re- 
move to  the  left  of  the  unit's  phice ;  hence  if  x  represent  a  digit  in  the 
place  of  tens,  and  1/  in  the  place  of  units,  the  number  will  be  expressed 
by  lOx -}-?/.  A  number  consisting  of  three  places,  with  z,  y,  and  2,  to 
represent  the  digits,  will  be  expressed  hy  IOO2;  -\-  10?/  -j-  z. 

Ans.   36. 

7.  A  number  is  expressed  by  three  figures ;  the  sum  of 
these  figures  is  9  ;  the  figure  in  the  place  of  units  is  double 
that  in  the  place  of  hundreds,  and  when  198  is  added  to  this 
number,  the  sum  obtained  is  expressed  by  the  figures  of  this 

^  number  reversed  ;  what  is  the  number  ?  Ans.  234. 

8.  Divide  the  number  90  into  three  parts,  such  that  twice 
the  first  part  increased  by  40,  three  times  the  second  part  in- 
creased by  20,  and  four  times  the  third  part  increased  by  10, 
may  be  all  equal  to  one  another. 

Ans.  First  part,  35;  second,  30  ;  and  third,  25. 

9.  Find  three  numbers,  such  that  the  first  with  ^  of  the 
other  two,  the  second  with  \  of  the  other  two,  or  the  third 
with  I  of  the  other  two,  shall  be  equal  to  25. 

Ans.  13,  17,  and  19. 


160  SIMPLE  EQUATIONS. 

10.  There  are  three  numbers,  such  tliat  the  first  with  |  the 
second,  is  equal  to  14  ;  the  second  with  ^  of  the  third,  is 
equal  to  18  ;  and  the  third  with  ^  of  the  first,  is  equal  to  20  ; 
required  the  numbers.  Ans.  8,  12,  and  18. 

11.  Find  three  members,  such  that  i  of  the  first,  4  of  the 
second,  and  ^  of  the  third  shall  be  equal  to  62  ;  ^  of  the  first, 
I  of  the  second,  and  |  of  the  third  equal  to  47  ;  and  -^  of  tie 
first,  I  of  the  second,  and  ^  of  the  third  equal  to  88. 

Ans.  24,  GO,  and  120. 

12.  Find  three  numbers  of  such  magnitude  that  the  first 
with  the  ^  sum  of  the  other  two,  the  second  with  |  of  the  other 
two,  and  the  third  with  ^  of  the  other  two,  may  be  the  same, 
and  amount  to  51  in  each  case.  Ans.  15,  33,  and  39. 

/\  13.  Four  boys,  A,  B,  C,  and  D,  comparing  their  money, 
found  that  A's  money  added  to  ^-  of  the  sum  possessed  by  the 
other  three,  would  make  $30 ;  B's  money  added  to  ^  of 
the  sum  possessed  by  the  other  three,  would  make  $32  ;  C's 
money  added  to  -3-  of  the  sum  possessed  by  the  other  three, 
would  make  $34  ;  and  D's  money  added  to  ^  of  the  sum  pos- 
sessed by  the  other  three,  would  make  $36  ;  what  sum  had 
each  ?  Ans,  A,  $12  ;  B,  $15  ;   C,  $18  ;   D,  $21. 

14.  The  sum  of  three  fractions  is  2.  The  second  fraction  is 
double  that  of  the  first ;  and  the  third  is  double  that  of  the 
second  ;  what  are  the  fractions  ?  Ajis.   |,  1,  and  |. 

15.  The  first  of  three  numbers  with  ^  of  the  other  two 
make  23  ;  the  second  with  ^  of  the  other  two  make  30  ;  the 
third  with  twice  the  sum  of  the  other  two  make  72.  What  are 
the  numbers?  Ans.  12,  15,  18. 

16.  A's  age  is  double  that  of  B's,  B's  is  triple  that  of  C's, 
and  the  sum  of  all  their  ages  is  140  ;  what  is  the  age  of  each  ? 

Ans.  A's  =  84;  B's  =  42  ;  C's  =  14. 

17.  A  man  WTOught  10  days  for  his  neiglihor,  liis  wife  4 
(lays,  and  their  son  3  days,  and  tliey  all  received  11  dollars  50 
cents ;  at  another  time  the  man  served  9  days,  his  wife  8  days, 


THREE  OR  MORE  UNKNOWN  QUANTITIES.  ^gi 

and  tlie  son  6  days,  at  the  same  rates  as  before,  and  rei^eived 
12  dollars ;  a  third  time  the  man  served  7  days,  his  wife  li 
days,  and  the  son  4  days,  at  the  same  rate&  as  before,  and  re- 
ceived iJ  dolhirs.     What  were  the  daily  wages  of  each  ? 
Ans,  Husband's  wages,  $1.00  ;  wife's,  0  ;  son's,  50  cents. 

NEGATIVE   RESULTS. 

1»>7.  Til  the  solution  of  the  last  example,  the  wages  of  the 
wife  are  found  to  be  0,  which  means  that  she  received  no 
wages.  The  following  examples  will  illustrate  negative  re- 
sults : 

1.  A  man  worked  for  a  person  10  days,  having  his  wife  with 
him  8  days  and  his  son  6  days,  and  he  received  10  dollars  30 
cents  as  compensation  for  all  three  ;  at  another  time  he  wrought 
12  days,  his  wife  10  days,  and  son  -t  days,  and  he  received  13 
dollars  and  20  cents;  and  at  another  time  he  wrought  15  days, 
his  wife  10  days,  and  his  son  12  days,  at  the  same  rates  as  be- 
fo'-e,  and  he  received  13  dollars  85  cents.  What  were  the 
daily  wages  of  each  ? 

Ans,  Husband,  75  cents  ;  w^fe,  50  cents;  son,  —  20  cents. 

The  sign  minus  signifies  the  opposite  to  the  sign  plus. 
Hence  the  son,  instead  of  receiving  wages,  was  at  an  e.rjjensc 
of  20  cents  a  day,  and  the  language  of  the  problem  is  thus 
shown  to  be  incorrect. 

2.  Two  men,  A  and  B,  commenced  trade  at  the  same  time ; 
A  had  3  times  as  much  money  as  B,  and  continuing  in  trade, 
A  gained  400  dollars,  and  B  150  dollars ;  A  then  had  twice 
as  much  money  as  B.     How  much  did  each  have  at  first  ? 

Without  any  special  consideration  of  the  problem,  it  implies 
that  both  had  money,  and  asks  how  much.  But  on  solving 
the  problem  with  x  to  represent  A's  money  and  y  B's,  we  find 

a;  =  —  300 
and  2/  =  —  100  dollars. 
14*  L 


162  SIMPLE   EQUATIONS. 

That  is,  they  had  no  money,  and  the  minus  sign  in  this  ease 
indicates  debt;  and  the  solution  not  only  reveals  the  numeri- 
cal values,  but  the  true  conditions  of  the  problem,  and  points 
out  the  necessary  corrections  of  language  to  correspond  to  an 
arithmetical  sense. 

The  problem  should  have  been  written  thus : 

A  was  three  times  as  much  in  debt  as  B ;  A  gains  400 
dollars,  and  13  150 ;  A  now  has  twice  as  much  money  as  B. 
How  much  were  each  in  debt  ? 

Ans,  A's  debt,  $300;  B's,  $100. 

These  results  are  positive,  and  show  that  the  enunciation 
corresponds  to  the  real  circumstances  of  the  case 

3.  What  number  is  that  whose  fourth  part  exceeds  its  third 
part  by  12  ?  Ans,  —  144. 

But  there  is  no  such  abstract  number  as  — 144,  and  we  can- 
not interpret  this  as  debt.  It  points  out  error  or  impossihility^ 
and  by  returning  to  the  problem  we  perceive  that  a  fourtb 
part  of  any  number  whatever  cannot  exceed  its  third  part ;  it 
must  be,  its  third  part  exceeds  its  fourth  part  by  12,  and  the 
enunciation  should  be  thus : 

What  number  is  that  whose  third  part  exceeds  its  fourth 
part  by  12,  An^.  144. 

Thus  do  equations  rectify  subordinate  errors,  and  point  out 
special  conditions. 

4.  A  man  when  he  was  married  was  30  years  old,  and  his 
wife  15.  How  many  years  must  elapse  before  his  age  will  be 
three  times  the  age  of  his  wi^e  ?  Ans,  — 7^  years. 

The  question  is  incorrectly  enunciated ;  7i  years  before  the 
marriage,  not  after,  their  ages  bore  the  specified  relation. 

5.  What  fraction  is  that  which  becomes  -|  when  1  is  add(3d 
to  its  numerator,  and  I  when  1  is  added  to  its  denominate]'. 

Ans.  In  an  arithmetical  sense,  there  is  no  such  fraction 
The  algebraic  expression,  zj|,  will  give  the  required  results. 


INVOLUTION.  163 


SECTION  III. 
INVOLUTION; 

OR,   THE    FOllMATION    OF    POWERS. 

I #18.  A  Power  is  the  product   obtained   by  repeating  a 
quantity  several  times  as  a  factor. 

1«>l>,  Powers  are  indicated  by  exponents,  from  which  thej 
take  their  names. 

Thus,  let  a  represent  any  quantity : 


Its  fir  Hi  power  is 

a=::z  a^ 

Its  second  power  is 

aa  =  a" 

The  third  power  is 

aaa  =  a' 

The /oi/WA  power  is 

aaaa  =  a^ 

The  fifth  power  is 

aaaaa  =  a' 

In  general  terras,  a  to  the  ??th  power  is  aa^  &c.,  to  n  factors, 
and  n  may  be  any  number  whatever. 

IGO.  The  First  Power  of  any  quantity  is  the  quantity 
itself. 

The  Square  of  any  quantity  is  its  second  power. 
The  Cube  of  any  quantity  is  its  third  power. 

101.  A  Perfect  Power  is  a  quantity  that  can  be  exactly 
produced  by  taking  some  other  quantity  a  certain  number  of 
times  as  a  ftictor;  thus,  a^  -{-■  2ab  +  ^^  is  a  perfect  power,  be- 
cause it  is  equal  to  (a  -f  6)  x  («  +  b)  ;  x^  -f-  o.i:^  +  8^  +  1 
is  a  perfect  power,  because  it  is  equal  to  (x  -f  1)  (a?  +  1)  (a^  4-  1). 

iv^oTE. — Tt  i«  thoufiht  host  to  omit  qne«tioTis  at  the  bottom  of  the  pngea 
in  the  remaining  pait  of  this  work  leaving  the  teacher  to  use  such  as 
may  be  deemed  appropriate. 


164  INVOLUTION. 

ICI^,  An  Imperfect  Power  is  a  quantity  that  cannot  be 
exactly  i)ro(Ju('ed  by  taking  another  quantity  any  number  of 
times  as  a  factor;   as,  «^  -\-  by  x  +  oy,  and  d^  ~{-  ab  +  b'\ 

IGJI.  Involution  is  the  process  of  raising  any  quantity  to 
any  given  power.  Involution,  in  algebra,  is  performed  by  suc- 
cessive multiplications,  as  in  arithmetic. 

The  first  power  is  the  quantity  itself. 

The  second  power  is  the  product  of  the  quantity  multiplied 
bj  itself. 

The  third  power  is  the  product  of  the  second  power  by  the 
quantity. 

The  fourth  powder  is  the  third  power  multiplied  by  the  quan- 
tity, etc. 

POWERS  OF  MONOMIALS. 

104.  In  the  power  of  a  monomial  there  are  three  things 
to  be  considered  :  1st,  the  coefficient ;  2d,  the  exponents  ;  ^d, 
the  sign. 

1st.  With  respect  to  the  coefficient: 

Let  it  be  required  to  raise  'la  to  the  third  power:  w^e  have 
2a  X  2a  x  2a  =  2  x  2  x  2a'  =  2V  =  8al 

Hence,  The  coefficient  may  be  raised  to  the  required  power 
separately. 

2d.  With  respect  to  the  exponents : 

We  observe  that 

The  second  power  of  aMs     a^  x  a^  =  a^"^^  =  a*. 

The  tliird  power  of  a^  is  a?  x  af  X  a^  =  a^^^^^  —  al 

The  nth  power  of  a'  is  a^  x  a^  x  a^  x  etc.  =  a^+^+'  ^etc. 

Hence,  Thp  exponent  is  repeated  as  many  times  as  there 
are  units  in  the  index  of  the  power. 

3d.  With  respect  to  the  law  of  signs  : 

It  is  obvious  tliat  the  re])etition  of  any  positive  quantity  as 
a  factor  must  produce  a  positive  result.  But  the  successive 
powers  of  negative  quantities  must  have  varying  signs. 


POWERS  OF  MONOMIALS.  \Q^ 

Lei  K  be  required  to  raise  —  a  to  successive  powers.     We 
have 

Second  power,  —  a   x  —  a  =  -f  a^  positive. 
Third  power,     +  a^  x  —  a  =  —  a"\  negative. 
Fourth  power,  —  a^  x  —  a  =  -f  a*,  positive. 
Fifth  power,       +  a*  X  —  a==  —  a\  negative. 
Ilence, 

1st.    All  the  powers  of  a  positive  quantity  are  positive, 
2d    The  even  powers  of  a  negative  quantity  are  positivey 
and  th4  odd  powers  negative. 

16tS.  From  these  principles  we  deduce  the  following 

Rule.     I.  Baise  the  numeral  coefficient  to  the  required 
power. 

II.  Multiply  the  exponent  of  each  letter  by  the  index  of 
the  required  power. 

III.  When  the  quantity  is  negative,  give  the  odd  powers  the 
minus  sign. 

EXAMPLES   FOR   PRACTICE. 

to  the    od  power.  Ans.  a^. 

to  the  4th  power.  Ans.  y^. 

to  the  5th  power.  Ans.  F^'\ 

to  the  4th  power.  Ans.  x^'\ 

to  the   8d  power.  Ans.  y^^. 

to  the  6th  power.  Ans.  x^. 

to  the mth  power.  Ans.  a:'"", 

to  the   8d  power.  Ans.  aV. 

9    Raise  ab'^x*  to  the    2d  power.  Ans.  a-7/j;^ 

10.  Raise  cV    to  the  5th  power.  Ans.  c^y^. 

11.  Required  the  od  power  of  Sax^.  Ans.  27aV. 

12.  Required  the  od  power  of  —  2.t.  Ans.  ^-8x1 

13.  Required  the  4th  power  of  —  B^r.  Ans.  Slx\ 

14.  Required  the  2d  power  of  Sa'b^.  Ans.  6Aa*b\ 

15.  Required  the  Bd  power  of  ^x^z  Ana.  12^a^z\ 


1. 

Raise  x^ 

2. 

Raise  y^ 

8. 

Raise  F' 

4. 

Raise  ^ 

5. 

Raise  y^ 

6. 

Raise  x'* 

7. 

Raise  x" 

8. 

Raise  ax"^ 

166  INVOLUTION.      . 

Expand  the  following  indicated  powers. 

16.  (  — 2a"7.  Ans.  —  32a^». 

17.  (^a'bcy,  Ans.  a'b'c^\ 

18.  {6ah/x)\  Ans.   2]Qa''y\t:^. 

19.  (2cr//c'y.         "  Ans.   IGaW'c'^ 

20.  (^8a'"6"')3.  Ans.  —  27g^"&^^ 
21  (Dni'''b')\    -—-.-,-_-  Ans,  24:3m^'"b^, 

22.  l  —  qr')\  Ans.  a"^. 

23.  (26-"*c-")^  Ans.  S^-'^^'c"**. 

POWERS  OF  FRACTIONS. 
166.  1.  What  is  the  3d  power  of  ^  ? 


OPERATION 


* 


/a \^^  a       a       a  a  x  a  x  a       o' 

Vc  /  ~"  c       c       c~'cxcxc~'<y^' 

Hence, "to  raise  fractions  to  powers,  we  have  the  following 
Rule.     Raise  both  numerator  and  denominator  to  the  re- 
quired  power. 

*  Suppose  we  were  required  to  raise  —  to  the  fifth  power,  and  did  not 

b 

know  whether  the  denominator  was  to  be  raised  or  not,  we  could  decide 

the  point  by  means  of  an  equation,  as  follows  : 

The  fraction  has  some  value,  which  we  represent  by  a  symbol,  say  P. 

Then  P-=    .     Now  if  we  can  find  the  true  5th  power  of  P,  it  will  be  the 
o 

required  5th  power  of  the  fraction. 

Clearing  the  equation  of  fractions,  we  have 

bP=a 

Taking  the  5th  power  of  both  members  giyes 

a* 
By  division,  P*  =  --. 

This  equation  shows  that  to  raise  any  fraction  to  any  power,  the  nume- 
rator and  denominator  must  be  raised  to  that  power. 


m 


POWERS  OF  BINOMIALS.  \Q^ 


EXAMPLES   FOR   PRACTICE. 

2.  Re({uired  the  2d  power  of  ^^^ — .  Ans.  -^-j-* 

2a  64a* 

3.  Required  the  Gth  power  of  —  ~.  Ans,  ^Tj-(i~~(r 

a^b  729a''6* 

4.  Required  the  6th  power  of  j,— .  Ans,   ^^ — . 

5.  Required  the  6th  power  of  |a'6.  Ans.  ^^a^%^, 

3  9 

6.  Required  the  2d  power  of  '-,  Ans.  \, 

Expand  the  following  indicated  powers. 


7. 

©■ 

8. 

(-1)' 

9. 

(|)" 

10. 

(%°)' 

11. 

(  'y. 

12. 

13 

Ans.     .-3. 

16;y* 
Ans.    .,  ,-    .. 

Ans.  ^. 

81. rV 

Ans.  -r--T'iii- 
2o()a^6^ 

14.  What  is  the  cube  of  — -^ ? 

^""'^  27^^ ^' 

15    What  IS  the  square  of ^-7  i 

4a'-f  4a64-4aa;  +  ^'  +  25a:  +  ar' 


£53  INVOLUTION. 

POWERS  OF  A  BINOMIAL. 

167.  The  Leadings  Letter,  Q,nantity,  or  Term,  is  the  one 

wliich  is  written  first  in  the  binomial ;  the  other  is  called  the 
second,  or  following  letter,  quantity,  or  term. 

The  process  of  expanding  the  higher  powers  of  a  binomial 
by  actual  multiplication  is  very  tedious,  and  hence  mathema- 
iieians  long  since  sought  to  discover  some  shorter  method, 
^hich  a  method  was  first  developed  by  Sir  Isaac  Newton,  and 
is  known  as 

NEWTON'S  BINOMIAL   THEOREM. 

168.  In  order  to  more  clearly  investigate  the  properties  of 
different  powers  of  a  binomial,  we  will  first  obtain  a  few  powers 
of  a  +  6  by  continued  multiplication  ;  thus, 

Let  a  +  bhe  raised  to  the  2d,  3d,  4th,  &e.,  powera, 
a-^-b 
a  +  b 


2d  power, 


3d  power, 


4th  power. 


a'+    ab 

ab  +b'     • 
a'+  2ab  +  6» 
a  +  b 
a«+  'Ia:'b  4-    ab^ 

a:'b  +  2ab' 

+  6' 

a'+  Sa'b  -f  Sa¥ 

+  6» 

a'+  ^a^b  +  Sd'W 
a'b  +  na'b' 

+     ab' 

+  Sab'  +  6* 

a*-f  4a'b  +  6a'b'  +  4a6*  +  6* 
a-\-b 

a'+  4a^b  +  Qa'b'+  ^a^b'-h    a¥ 

a'b  +  4a'b'+  ea'b'+  4a6*+i^ 

5th  power,       a'+  ba'b  +10a-'6=^+10a'6^-f  5a6*-f  ^ 


POWERS   OF   BINOMIALS.  j^gy 

If  a  —  6  be  raised  to  the  same  powers,  we  have  i 


a  —  5 
a  —  b 
a^ — ab 
—  ab  +  b' 
2(1  power,        a^ —  2ab  +  b"^ 
a  —  b 


a'—2a'b-}-    ab' 
—   a'b  +  2ab'^b^ 


8d  power,        a^—  Sa'b  +  Sab'  —  b^ 
a  —  b 


a'—Sa'b  +  Sa'b'-^   ab' 
—   a'b  +  Sa'b'—  Sa¥  +  ¥ 
4th  power,       a* —  4a-^6  -f  Qa-b^ —  4a6^  +  6* 
a  —  b 


a^—  4a'b  +  6aW-^  ia;'b'+     ab' 
—   a'b  +  4a'¥-^Qa:'b'+  4:ab'—-b^ 
5th  jpower,       a'—  ba'b  +10a^62— 10a^6^+5a6*—  b^ 

By  inspecting  these  results  we  may  arrive  at  general  prin- 
ciples, according  to  which  any  power  of  a  binomial  may  be 
expressed,  without  the  labor  of  actual  multiplication.  In 
order  to  do  this  it  is  obvious  that  there  are  five  things  to  be 
considered : 

IsL  The  .  number  of  terms ;  2c?,  The  signs  of  the  terms ; 
8(i,  The  letters ;  Uh^  The  exponents ;  bth^  The  coefficients 

Ist,  The  number  of  terms 

1^9,  We  observe  that  in  the  second  power  there  are  three 
terms  ;  in  the  third  power  there  are  four  terms  ;  in  the  fourth 
power,  five  terms  ;  and  in  the  fifth  power,  six  terms.     Hence, 

The  number  of  terms  is  always  greater  by  one  than  the 
index  of  the  power. 


170  INVOLUTION. 

2d,  The  signs  of  the  terms : 

170.  We  see  that  all  the  terms  iu  the  powers  of  a  f  h  are 
positive ;  but  in  the  powers  of  a  —  6,  the  signs  plus  and 
minus  alternate,  the  first  term  being  positive,  the  second  nega- 
tive, and  so  on.     Hence, 

I.  If  both  terms  of  the  binomial  have  the  plus  sign^  all  the 
tei^ms  of  any  power  will  be  positive. 

II.  But  if  the  second  term  of  the  binomial  have  the  minus 
sign,  all  the  odd  terms,  counting  from  the  left^  will  be  posi- 
tive, and  all  the  even  terms  negative. 

M.  The  letters: 

171.  By  inspecting  any  power,  we  perceive  that, 

The  second  letter  or  quantity  does  not  appear  in  the  first 
term.;  the  leading  letter  or  quantity  does  not  appear  in  the 
last  term;  and  both  letters  or  quantities  appear  in  all  the 
intermediate  terms. 

4:th.  The  exponents : 

172.  We  observe  that  in  the  fifth  power  of  both  binomials, 
the  exponents  of  the  letters  in  the  several  terms  are  related  as 
follows,  from  the  first  to  the  last : 

Of  the  leading  letter,  ,5        4        3  2  1 

Of  the  second  letter,  12  3  4        5 

Sum,  5        6        5  5  5        5 
Hence, 

I.  The  exponents  of  the  leading  letter  or  quantity  m  the 
successive  terms  form  a  series,  commencing  in  the  fii^st  term 
with  the  index  of  the  power,  and  diminishing  by  1  to  the 
right. 

IT.  The  exponents  of  the  second  letter  or  quantity  form  a 
series  commencing  in  the  second  term  with  1,  and  increasing 


POWERS  OF  BINOMIALS.  \71 

by  1  to  the  last  term,  in  which  the  exponent  is  equal  to  the 
index  of  the  poicer.  • 

III.  The  sum  of  the  exponents  iri  any  term  is  equal  to  the 
index  of  the  power. 

5th    The  coefficients : 

1*73.  The  law  governing  the  coefficients,  though  less  obvi- 
ous, may  be  exhibited  as  follows,  taking  the  5th  power : 

1st  term  is  la'\  and  1  X  5  =  5,  coefficient  for  2d  term  ; 

5x4 

2d  term  is  5a^6*  and  — ^ —  =  10,  coefficient  for  3d  term ; 

3d  term  is  10a^b'\  and  ^ — ^^ —  =  10,  coefficient  for  4th 
term : 

]0  X  2 
4th  term  is  lOa'6^,  and  - — j —  =  5,  coefficient  for  5th  term ; 

5x1 
5th  term  is  5a6*,  and  — ^ —  =  1,  coefficient  for  6th  term ; 

6th  term  is  W. 

Hence, 

1.  The  coefficient  of  the  first  term  is  1. 

II.  The  coefficient  of  the  second  term  is  the  index  of  the 
required  power. 

III.  The  coefficient  of  any  term  multiplied  by  the  expo- 
nent of  the  leading  letter  or  quantity,  and  divided  by  the  ex- 
ponent of  the  second  letter  or  quantity  plus  1,  will  be  the 
coefficient  of  the  next  succeeding  term. 

Notes.  1.  It  will  be  SQen  that  the  coefficients  of  the  last  half  of  the 
terms  are  the  same  as  the  coefficients  of  the  first  half  inversely,  and  that 
the  coefficients  of  anj'-two  terms  at  equal  distances  from  the  extremes  are 
equal.     Hence  the  labor  of  computing  them  may  be  avoided. 

2.  In  obtaining  the  coefficient  of  any  term,  there  are  two  operations, 
multiplication  and  division,  and  cancellation  can  always  be  applied. 

3.  The  exponent  of  the  second  letter  or  quantity,  plus  1,  is  alwaya 
equal  to  the  number  of  the  term,  counting  from  the  left. 

Note. — We  have  now  established  by  induction,  and  observations  upou 
particular  cases,  the  principles  of  Newton's  Theorem.  Its  rigid  demon- 
stration is  somewhat  difficult,  but  its  application  is  simple  and  practical. 


172  INVOLUTION. 

EXAMPLES   FOR  PRACTICE. 

1.  Expand  (^  -1-  yy,  Ans.  x^  +  ^dx^y  +  ^xa/   \-  y^, 

'J.  Expand  {y  +  z)^, 

Ans.  y'  +  ly^z  +  21  i/V  +  35i/V  +  So^/V  +  2lyh^ -j-  Ixjz^  -^  z  . 

3.  Expand  {a  +  h)\ 

Ans.  a«  +  ^a'h  +  28a^'62 f  SCa^ft^  +  70a^6*  +  56a»6^  -f- 
28a266+8a6^+6l 

4.  Expand  (a  —  6)*. 

J[ns.  a*  — 4a'^6  +  Ga^ft*  — 4a6^  +  6* 

5.  Expand  (u:^  +  yy. 

Ans.  x^  4-  'ox^y  +  15a;y  +  20a;y  +  15a;y  +  G:?^/'^  -f  y^ 

6.  Expand  (x  —  yy, 

Ans.  x^—6x^y  +  ISa^y— 20xY+  15xy  —  Gxi/^  +  y». 

7.  Expand  (a  +  6)^^. 

^??s.  ai«+10a^6+45a^62^120a^6-^  +  210a«6^4-  252a56M- 
210a^6«  4-  120aW  +  45a'^6«  +  10a6'^  +  b'\ 

8.  Expand  (ac  +  xy. 

Ans.   a^d^  +  ^Ou'c^x  +  Ca'^cV  +  4ac.a:;^  +  x^. 

9.  Expand  (c  +  ax^^. 

Ans.  &  +  h&'ax  +  lOc^aV  +  10cV.r^  +  5caV  +  a^^.-*. 

10.  Expand  (ah  +  c.r?/)^ 

^7?s.  ah  +  oa6  c^2/  +  '^cthcxy  +  cr?/ . 

11.  Expand  (a  +  1)1 

Note. — Notice  that  all  powers  of  1  are  1,  -wliich  is  not  written  when  a 
factor;  and  that  the  divisor  in  obtaining  the  coefficients  will  be  the  num- 
Iwr  of  the  term  employed  counted  from  the  left. 

Ans.  a^'+  5a* -f  lOa'  4-  10a''  +  5a  +  1. 

12.  Expand  (1  —  a)^ 

Ans.  1  —  5a  +  lOa'^  —  lOa^  +  5a*  —  a\ 

13.  Expand  (z  —  1)^ 

Ans.  z'-^Gz'  +  15z'  —  20z^  +  152^^  — 6z  H-  1. 

14.  Required  the  third  power  of  Sx  +  2y. 


POWERS  OF  BiNOM|ALS.  f  173 

We  cabinet  well  expand  this  by  tl|e  binomial  theorem,  be- 
cause the  terms  are  not  sii^nple  liiercl  qiioMities.  But  we  can 
assume  ox  =-  a  and  2y/  =;:;  5.     The;/  ; 

8^  +  2y  =  a  +  b,  and  (a  -f'  6)^  ^  a^  +  Ua'b  +  Sab'  +  5?. 

Kow  restoring  the  values  of  a  and  6,  we  have, 

Sa'b  =  S  X  dx'  x22j  =  Mx'y 
Zab''  ~  8  X  oa;   X  4y2  ==  ZQxif 
b'  ==  Si/' 

Hence,  (Sx  +  2yf  =  21  x^  +  b^xhj  +  oG^^^  ^  3^^ 

15.  Required  the  4th  power  of  2a^  —  3. 

Let  X  =  2a^  ?/  =  3.  Then  expand  {x  —  y)\  and  restore 
the  values  of  x  and  y,  and  the  result  will  be, 

16a«  —  96a«  +  -l^)a'  —  216a2  +  81, 

16.  Required  the  cube  of  {a  +  b  -{-  c  -\-d). 

As  we  can  operate  in  this  summary  manner  only  on  hino- 
mial  quantities,  we  represent  a  +  6  by  x,  or  assume  j?  =  a  +  Z>, 
and  y  =zG  -^  d.  / 

Then  (x  +  yf  =  x'  +  Sxhj  +  Sxy'  +  7/. 

Restoring  the  values  of  x  and  i/,  we  have 

(a  +  by  +  3(a  +  by(c  +  d^  +  3(a  -f  6)  (c  +  dy  +  (c  +  (Q«. 

Now  we  can  expand  the  binomial  quantities  contained  in 
parentheses. 

The  method  of  substitution  which  we  have  been  obliged  to 
adopt  in  the  last  three  examples,  has  been  long  in  use  among 
mathematicians  for  expanding  binomials  with  coefficients. 

174.  A  direct  method  of  obtaining  the  coefficients  in  the 
expansion  of  a  bhiomial  in  which  the  terms  have  coefficients, 
has  been  developed  by  J.  II.  French.  LL.U.,  Superintendent 
of  the  New  York  State  Map  and  Atlas  Survey,  and  we  present 
it  under  the  name  of 
15* 


J74  INVOLUTION. 

FRENCH'S   THEOREM. 

Any  binomial  having  coefficients,  may  be  involved  by  actual 
multiplication. 

llequired  the  5th  power  of  2a  +  ox. 
2a  +  ox 
2a  +  3x 
•2d  power,     4d'+12ax     +  dx^ 
2a+Sx 
SaF+2Aa'x     +  ISax^ 

V2a'x     4-  36aa;^    -f  27.?^ 
3d  power,     Sa'-i-SQa^x     -f-  54aj;^    4-*-^7x-^ 
2a  '■\-ox 
iQa'+f2a'x     +108aV  +  ^^:ax' 

24:a'x     -f  lOSa'V  +I(j2ax'  +  81  j;* 
4ili  power,  16a'+9Qa^x     +216a-;rH^T6a5^~4^T^ 
2a  4-  o-r. 
82a^  +  192a*^+432aV  +432aV4-102a^ 

48a*^+ 288aV+648aV-f-648a.:r*+248^ 
5th  power,"32^^+240a*j;4-T20aV+l080aV+810aj;*+243j;» 

ITo,  By  a  close  analysis  of  the  result,  we  may  arrive  at 
general  principles  which  will  enable  us  to  expand  any  binomial 
having  coefficients  with  the  same  facility  that  Newton's  Theo- 
rem enables  us  to  expand  to  any  power  a  -\-  x.  If  we  examine 
the  result  in  the  same  order  as  we  did  in  (IG8-I7S^),  we  shall 
find  that  there  is  no  difference  in  the  expanded  form  of  (a  -j-  a;),* 
and  (2a  +  3^)^,  except  in  the  coefficients.     That  is, 

In  any  power  of  a  binomial,  the  number  of  terms,  the  signs, 
the  letters,  and  the  exponents  of  the  literal  part  in  the  several 
terms  are  independent  of  the  coefficients  of  the  binomial  root 

We  will  therefore  confine  our  analysis  to 

The  Coefficients. 

ITG.  The  rigid  demonstration  of  the  law  which  governs  tne 
formation  of  the  coefficients  being  too  difficult  for  this  place, 
Is  reserved  for  the  University  Algebra.  Bnt  tlie  law  itself  may 
be  exhibited  as  follows  : 


POWERS  OP  BINOMIALS.  I75 

1st  coefficient,    32  =  2^,  i.  e.   the  coefficient  of  the  leading 

quantity  in  the  root  raised 
*  to  the  power  of  the  given 

index. 

82  X  5  X  3 
2d  coef.,  240  = ,  i.  e.  the  product  of  the  co- 
efficient of  the  first  term,  the 
exponent  of  the  leading 
quantity  in  the  first  term, 
and  the  coefficient  of  the 
following  quantity  in  the 
root,  divided  by  the  coeffi- 
cient of  the  leading  quantity 
in  the  root. 

o ,       c\ocx        240  X  4  X  8      .         .  ^    .    ^  ., 

od  coef.      720  == ^ -^ ,  i.  e.  the  product  of  the  co- 

»  efficient  of  the  second  term, 
the  exponent  of  a  in  that 
term,  and  the  coefficient  of 
X  in  the  root,  divided  by  the 
product  of  tlie  coefficient  of 
a  in  the  root  and  the  num- 
ber of  the  term,  counting 
from  the  left. 

4th  coef.,  1080  =  —^-f^ — ,  ^'.  ^-  the  product  of  the  co- 
efficient of  tlie  third  term, 
the  exponent  of  a  in  that 
term,  and  the  coefficient  of 
X  in  the  root,  divided  by 
the  product  of  the  coeffi- 
"^^-.,_.^  cient  of  a  in  the  root  and 

the  number  of  the  term. 
(Or,  which  is  the  same 
thing,  the  exponent  of  x 
m  the  third  term  -f  1  ) 


176  INVOLUTION.     ' 

1080  >f  2  y   '^ 
6th  coef.,  810  = — -^  i.  e.  the  product  of  the  co 

efficient  cf  the  fourth  te]"]u, 
the  exponent  of  a  in  that 
term,  and  the  coefficient  of 
X  in  the  root,  divided  by  the 
product  of  the  coefficient  of 
a  in  the  root  and  the  num- 
ber of  the  term. 

6th  eoef.,  243  = ^T^TK >  ^'-  ^-  ^^^^  product  of  the  co- 
efficient of  the  fifth  term, 
the  exponent  of  a  m  that 
term,  and  the  coefficient  of 
X  in  the  root,  divided  by 
the  product  of  the  coeffi- 
cient of  a  in  the  root  and 
the  number  of  the  term. 

177.  From  this  example  and  analysis  we  may  deduce  the 

LAW   or   THE    COEFFICIENTS. 

I.  The  coefficient  of  the  first  term  in  any  power  is  ahvoi/a 
equal  to  the  corresponding  power  of  the  coefficient  of  the 
leading  term  in  the  root, 

II.  The  coefficient  of  the  second  tei^m  is  obtained  by  mnU 
tip)lying  the  first  coefficient  by  the  exponent  of  the  leading 
quantity,  and  this  jjroduct  by  the  coefficient  of  the  following 
quantity  in  the  root,  and  dividing  by  the  coefficient  of  the 
leading  quantity  in  the  root. 

Universally; —  The  coefficient  of  any  term  may  be  ob- 
tained hy  multiplying  the  coefficient  of  the  preceding  term  by 
the  exponent  of  the  leading  quantity  in  that  term,  or  by  the 
number  of  the  term  from  the  last,  and  by  the  coefficient  of 
the  following  quantity  in  the  root,  and  dividing  this  i^esultby 
the  product  of  the  coefficient  of  the  leading  quantity  in  the 
root,  multiplied  hy  the  number  of  the  term  from  the  first. 


POWERS  or  BINOMIALS.  J[77 

Note. — Tbe  coefficient  of  the  last  term  in  any  power  is  alw&ys  equal 
to  the  corresponding  power  of  the  coefficient  of  the  following  term  in  the 
root. 

If  8.  There  is  another  class  of  biuomials  that  come  under 
a  modification  of  this  Theorem,  viz.  :  those  having  exponents. 
To  illustrate  the  application,  let  us  write  out  the  fourth  power 
of  2a'  +  SxK 

2a^  +  Sx^ 

2d  power,     4a«  +  12a\x^  +    9x* 
2a»  -f     Sx' 

8a»  +  2iaV  +     ISa^x* 

12a'x^  +     86aV  +  27xS 

8d  power,     8a»  -f-  oQa^^^  +    b4:a'x\  +  27x« 
2a^  +    ^x' 

l(ia''+  72aV  +  lOSa^^*  +     54aV 

24aV  +  lOSa^x'  +  162a»x«  +  81x» 

4th  power,  16ai2+  95^,9^2  j_  2l6a'x'  +  216aV  +  81^ 

On  examining  the  result,  we  shall  find  that  the  exponent  of 
the  leading  letter  is  12(=="*  ^  ^)  in  the  first  term,  and  that  it  di- 
minishes regularly  by  3  in  each  succeeding  term.  Also  that 
the  exponent  of  the  following  letter  is  2  in  the  second  term, 
and  increases  regularly  by  2,  in  each  succeeding  term,  to  the 
last,  where  it  is  8(=-'^^).  Hence  the  exponents  are  governed 
by  the  law  of  Newton's  Theorem,  as  shown  in  (17^),  modi- 
fied by  the  values  of  the  exponents. 

The  coefficients  are  the  same  as  the  coefficients  of  (2a  +  S:v)\ 
(174),  and  may  be  obtained  in  the  same  manner,  if  we  keep 
constantly  in  mind  tlie  fact  that  the  first  exponent,  12,  is  the 
exponent  3  of  the  leading  quantity  in  the  root  raised  to  the 
fourth  power,  and  that  the  real  exponent  which  we  are  to  use  as 
a  factor  of  our  dividend  is  the  exponent  of  the  leading  quantity 
in  any  term  divided  by  the  exponent  of  tlie  leading  quantity 

M 


178  INVOLUTION. 

in  the  root.  But,  as  this  is  liable  to  be  forgotten,  we  can  nse 
the  exponent  ol  the  leading  quantity  in  any  term,  whatever  it 
may  be,  as  a  factor  of  the  dividend,  if  we  write  the  exponent 
of  the  leading  quantity  in  the  root,  as  a  factor  of  the  divisor. 
Observing  this  direction,  and  the  indicated  operations  for  ob- 
taining the  several  coefficients  in  (2a^  -f-  ox^y^  will  be  as 
follows  : 

1st  coefficient,  2*  =  16 

2d  coefficient,        ^ — '^ — -  =  96 

3x2 

8d  coefficient,  ^  ^  g  ^  ^    =  216 

A,x.        ^  '    ,      216  X  6  X  3        ._ 

4th  coefficient,     -x ^ ^r-  =216 

3x3x2 

p;.!.        «,  .     ,      216  X  3  X  3        _^ 

oth  coefficient,     —. ?r — — -  =81 

'       4x3x2 

S79.  Examining  the  indicated  operations  for  obtaining  the 
coefficients  of  the  expanded  form  of  (2a  +  Sx^  (177),  we 
observe  the  following  facts  : 

IsL  Each  dividend  after  the  first  term  is  composed  of  three 
factors,  the  first  of  which  is  the  coefficient  of  the  preceding 
term,  the  second,  the  exponent  of  the  leadiug  quantity  in  the 
preceding  term,  and  the  third,  the  coefficient  of  the  following 
quantity  in  the  root. 

2d.  Each  divisor  is  composed  of  two  factors,  the  first  of 
which  is  the  number  of  the  preceding  term  counted  from  the 
left,  and  the  second  the  coefficient  of  the  leading  quantity  in 
the  root. 

3cZ.  The  second  factor  of  the  dividend  decreases  regularly 
by  1,  and  the  first  factor  of  the  divisor  increases  regularly 
by  1,  in  each  succeeding  coefficient. 

^tJi.  The  third  factor  of  the  dividend,  and  the  second  factor 
of  the  divisor,  are  the  same  in  each  coefficient,  i.  e.,  they  are 
constant. 


POWERS   OP  BINOMIALS.  X79 

We  may,  therefore, 

Let  a=  1st  coefficient  of  any  binomial. 

6  =  2d  coefficient  of  any  binomial. 

n  ==  the  exponent  of  any  binomial. 

Then      {ax±z  hyy  any  power  of  any  binomial. 

j^ssume  the  first  coefficient  to  be  Ci,  the  second  C2,  the 

third,  C3,  &c.,  and  we  shall  have  the  following 

General  Formula  for  Coefficients. 
Ci  =  a'^ 
_  Qvnh 
a 


C3  ==  ^'^^^-^)^ 


C 


'la 

Q,(n  —  2)^ 

oa 
Q^(n  —  3)6 


4a 


&c.  &c. 


Note.  We  have  now  carried  the  investigation  of  this  Theorem  as  far  as 
the  plan  and  limits  of  this  work  admit.  It  is  general  in  its  application,  and 
may  be  used  in  the  involution  of  ^i\y  binomial  whatever.  Its  full  develop- 
ment, including  its  application  to  negative  indices  and  binomial  roots,  will 
be  found  in  future  editions  of  the  University  Algebra. 

EXAMPLES   FOR   PRACTICE. 

1.  Required  the  4th  power  of  2a  +  8x. 

Ans.  16a*  +  96a^^  +  216aV  +  216aa;^  +  81x*. 

2.  Expand  (2a  —  bh)\ 

Ans.  8a'  —  60a^6  +  ISOaft^  —  125^8^ 
8,  What  is  the  cube  of  Ix  +  2ay  ? 

Ans.  843a;s  +  lUx'ay  +  84a:ay  +  8ay. 

4.  What  is  the  fifth  power  of  5a — 2c? 

Ans,  3125a^— .6250a*c4-5000aV— 2000aV+400ac*— 32e\ 

5.  Expand  (a;'  4-  3 2/^)5. 

Ans.  ^i»  +  \f)J^y''  +  90xy  4-  270xy  -f-  405:ry  +  243^1°. 


180  INVOLUTION. 

6.  Expand  (2a^  +  axf. 

Am.  8a^  +  I2a^x  +  6a'x'  +  2a'x\ 

7.  Expand  (x  —  !)«. 

Arts,  x'  —  Qx'  +  15x'^  —  20^8  +  15^=^  —  Qx-x-l. 

8.  Expand  (3a;  —  5)^ 

J.ns    21  x"  —  135x'^  +  225a"  — ■  125, 

9.  Expand  (Aa'b  -—  2c'y, 

Ans.  2^QaW  —  bl2aWc'  +  'SS4:a%'c'  —  128a^6e«  +  16aj . 

10.  Expand  (|  +  ^|)'. 

Note. — The  quantity  (^-\-  -~)  =  (  —  "  +  j ^)  • 

1    '  ^  15  4    _L  90    ,  ,   ,   270  ,  ,      405     ^   ,    243    . 

11.  Expand  (3  —  2r)  to  the  6tli  power. 

Ans,  729— 2916r+4860r^^4320r'+2160r*— 576r5  +  64rfi..     . 

12.  Expand  (x  +  —  j  to  the  7th  power. 
5 


13.  Expand  (l  +  o^) 


^        ,25         125  ,      625  3  .  8125^  ,  3125^ 
Ans.  l  +  -^x  +  —x^  +  -^x""  +  -jQ-^  +  -g2"^- 

3       5 


14.  Expand  (^ — ^xj 


81       45        75  ^      250  3      625 
Ans.  ^  —  —x^-^x ^x  +-gj-^. 

15.  Expand  (x'  —  Sy^. 

Ans.  ^i<'— 15xY+90a;y— 270^y+405^y— 243y^o. 


ROOTS  OP  MONOMIALS.  j81 


EVOLUTION; 

OR,  THE  EXTRACTION  OF  ROOTS. 


■^^ 


1§0.  A  Eoot  is  a  factor  repeated  to  form  a  power ;  or,  it 
is  one  of  the  equal  factors  of  a  quantity. 

181.  Evolution  is  the  process  of  extracting  tlie  root  of  a 
quantity.  It  is  tlie  converse  of  involution,  and  is  indicated 
by  the  radical  sign,  v^- 

18^.  The  Index  of  the  root  is  the  figure  placed  above  the 
radical  sign  to  denote  what  root  of  the  quantity  unde4-  the 
radical  is  to  be  taken  ;  thus,  in  i/a,  5  is  the  index  of  the  root, 
and  denotes  that  the  fifth  root  of  a  is  to  be  taivcn. 

183.  A  Surd  is  the  indicated  root  of  an  imperfect  power ; 
the  root  thus  indicated  cannot  be  exactly  obtained  or  ex- 
pressed ;  thus,  v/2  is  a  surd,  because  the  number  2,  not  being 
a  perfect  square,  can  have  no  exact  square  root.  A  surd  is 
sometimes  called  an  irrational  quantity. 

ROOTS  OF  MONOMIALS. 

18-4.  To  discover  the  process  of  extracting  roots,  we  must 
observe  how  powers  are  formed,  and  then  trace  the  operations 
bach.  Thus,  to  square  a,  we  double  its  exponent,  which 
makes  d\  (^7.)  The  square  of  a^  is  a^,  the  cube  of  a^  is 
a^,  &c.  The  4th  power  of  x,  is  x^ ;  the  ?ilh  power  of  x^  is 
^*";  &c.,  &c. 

Now,  since  multiplying  exponents  raises  simple  literal  quan' 
tities  to  j^owers,  dimdiiig  exponents  must  extract  roots.  Thus, 
the  square  root  of  a*  is  a^^  =  a^ ;  the  cube  root  of  a^  is 
a'^'  =  a\ 

The  square  root  of  a  must  have  its  exponent  (1  understood), 

divided  by  2,  which  will  give  a^  ;  the  cube  root  of  a  in  the 

like  manner  is  a"^,  and  the  exponents,  I,  ^,  I,  ^,  &c.,  indicate  the 
16 


X82  EVOLUTION. 

second,  third,  fourth,  and  fifth  roots  of  any  quantity  whose  ex- 

5 

poneut  is  1.  The  6th  root  of  ^  is  x^.  In  l!ke  manner,  |  ex- 
presses  the  4th  root  of  the  3d  power  of  a  quantity.  -  Hence 
the  foUowing  principles : 

I.  Boots  are  projjei^ly  expi^essed  by  fractional  exponents. 

II.  The  numerator  shows  the  power  of  the  quantity,  whose 
root  ts  to  be  extracted. 

III.  TJie  denominator  shoics  what  root  of  that  power  is  to 
he  extracted.     It  is  the  index  of  the  root. 

We  have  seen  (1114:)  that  any  power  of  a  positive  quantity 
is  positive,  and  tliat  the  even  powers  of  a  negative  quantity 
are  positive,  and  the  odd  powers  negative.  From  this  it 
results,  that 

I.  lire  odd  roots  of  a  positive  quantify  are  always  p)Osi- 
tive,  and  the  even  roots  are  either  jjositive  or  negative. 

II.  The  odd  roots  of  a  negative  quantity  are  negative,  and 
the  even  roots  are  impossible  or  imaginary. 

Note. — An  Imaginary  Quantliy  ia  the  indicated  even  root  of  a  negative 
quantity,  as  ^  IT^ or  ^Zl2^. 

1.  What  is  the  square  root  of  64a* j;^  ? 

OPERATION.  Analysis.     Since     the 

power   of  a  monomial  is 

(64aV)^±=  d=  8a'j?  formed  by  involving  each 

factor,  (165),  conversely, 

VERIFICATION.  *^°  ^""^  "^^^  ^^  "^'^'"^^ 

by  extracting  the  root  of 
(+  8a'^)  X  (+  ^a^x)  =  64a V  each     factor     separately. 

( —  %d'x)  X  ( —  Sa'^a;)  =  Qtla^x"^  The  square  root  of  6 1  is 

8  ;  of  a^  is  a^ ;  of  x^  is  x ; 
and  the  entire  root  is  Sa^ic,  to  which  we  give  the  double  sign,  zfc,  (re a 
plus  or  minus),  because  either  +  "^a^x,  or  —  %a^x,  squared,  will  pro- 
duce G4aAf^  as  is  seen  in  the  verification. 

185.  From  these  principles  and  illustrations  we  hare  the 
following 


ROOTS  OF  MONOMIALS.  1^3 

Rule.  I.  Extract  the  required  7^oot  of  the  numeral 
coefficient. 

II.  Divide  the  exponent  of  each  letter  by  the  index  of  the 
root. 

III.  Prefix  the  double  sign,  rb,  to  all  even  roots,  and  the 
minus  sign  to  the  odd  roots  of  a  negative  quantity. 

Note  1.  Under  this  rule  for  monomials  we  shall  introduce  no  numeral 
coefficient  the  required  root  of  which  will  consist  of  more  than  one 
place ;  hence  the  root  may  be  found  by  trial. 

EXAMPLES   FOR   PRACTICE. 

1.  What  is  the  second  root  of  ^a'^x^if  ?    Ans,  d=  Sax^y^, 

2.  What  is  the  third  root  of  Say  ?  Ans.  20% 

3.  What  is  the  fourth  root  of  81a*:c^^  ?       Ans.  d=  oax^ 

4.  What  is  the  fi:fth  root  of  32a^x^y^  ?         Ans.  2ax'f, 

5.  What  is  i\\Q  fourth  root  of  81a*Z)V^  ? 

Ans,  rtSafe^c^ 

6.  What  is  the  third  root  of  —  27a^  V  ? 

Ans,  —  3a*a;. 
Find  the  following  indicated  roots  : 

7.  {—21a^by,  Ans.  —  3«.^6, 

8.  (25xY)i  Ans.  ±  5xV. 

9.  v^lGa;*'".  Ans,   =fc  2x"'. 
10.  -v/oy.  Ans.  a^'if. 


11.  V125a«m».  Am,  Sa^wi^  or  ^a^^^, 

12    x/^  -4/15.  a^ 

18.  {xTy'^Y.  Ans,  xy^,  or  arV^. 

L  !!' 

14.  (a^6'0".  Ans,  a%  or  6y^ 

15.  Find  the  cube  root  of  4a*. 

Note  2    If  the  coefficient  is   an  imperfect  power,  it  may  be  treated 
as  a  literal  factor,  and  its  root  indicated. 

Ans,  4  V,  or  a^^f. 


184  EVOLUTICN. 

16.  Find  the  5th  root  of  7a^6^^  Ans.  l^Jb'. 

17.  Find  the  9th  root  of  —  lox^y^,       Ans,  —  15^x V'^- 

18.  Extract  the  square  root  of  — -. 

Note  3.  Since  the  power  of  a.  traction  is  formed  by  involving  the  nume- 
rator and  denominator  separately,  the  root  of  a  fraction  will  be  obtained 
by  extracting  the  root  of*the  numerator  and  denominator  separately. 


19.  Extract  the  cube  root  of  —  tt-t-.- 


7i\ 

20.  Extract  the  fourth  root  of  — . 


64  a^ 
21.  Extract  the  square  root  of  ^-z — ,. 


SQUARE   ROOT   OF   TOLYNOMIALS. 


Ans. 

3x 

[ns.  - 

""  2a  V 

A71S. 

1 

n ' 

Ans. 

9x 

186.  In  order  to  discover  the  process  of  extracting  the 
sqtiare  root  of  a  polynomial,  we  must  observe  how  the  squares 
of  polynomials  are  formed.    If  we  square  a  -{  b,  we  shall  ha^  e 

(a  +   by  =  a^  -f  2ab  4-  b\ 
This  result,  expressed  in  words,  is  as  follows  : 

The  square  of  (he  first  terii,  plus  twice  the  product  of  the 
two  terms,  plus  the  square  of  the  second  term. 

The  last  two  terms  of  the  power  may  be  factored  as  follows; 

2ab  -f  b'  =  (2a  +  b)b, 

which  is  expressed  thus  : 

Twice  the  first  term  plus  the  second,  multiplied  by  the 
second. 


SQUARE  ROOT  OF  POLYNOMIALS.  285 

1.  Extract  the  square  root  of  a'^  _|.  2ah  +  61 

OPERATION.  Analysis.     Reversing;  the  pro- 

^2  j^  2ah  +  &H  a  +  &  ^^^^  of  invclutlon,  we  extract  tiio 

2        ~  square  root  of  a\  and  obtain  a. 

^  the  first  term  of  the  root.     The 


2a  ^  h  )  2a6  +  6^  next  term  of  the  power  is  2ah  =: 

t;,   ,         ,  2  2a  X  &>  or  if^^7^■ce  the  first  term  cf 

tlie  root  multiplied  hy  ilie  second , 

we  therefore   divide  this  terra  hj 

2a,  twice  the  first  term  of  the  root,  to  obtain  6,  the  second  term  of 

the  root.     Placing  h  in  the  divisor  also,  at  the  right  of  2a,  we  have 

2a  +  h,  or  twice  the  first  term  plus  the  second,  which,  multiplied  by 

h,  gives  2a6  +  b\  the  last  two  terms  of  the  power. 

Again,  let  us  form  the  square  of  any  polynomial,  as  a  -f  6  +  o, 
in  the  following  nmnner  : 

Assume  s  =  a  -\-  b,  the  first  part. 

c  =  the  second  part. 

Then  (5  +  c)'  =  s'  +  2sc  +  cK     Hence, 

The  square  of  any  polynomial,  considered  in  itco  pjarts,  is 
equal  to  the  square  of  the  first  part,  plus  tivice  the  p)rodiict 
of  the  two  parts,  plus  the  square  of  the  second  p)art. 

Thus  the  root  of  any  quantity  can  be  brought  into  a  bino- 
mial, and  the  rule  for  a  binomial  root  will  answer  for  a  root 
containing  any  number  of  terms,  by  considering  the  root 
already  founds  however  great,  as  one  term,  or  one  part. 

2.  Find  the  square  root  of  a^  -j-  2ab  -f  ¥  -r^ac  +  2bc  -f  c*, 

OPERATION. 

a''  +  2ab  +  b' +  2ac  +  2bc  +  o\a  +  b-hc) 
a' 


la  +  b         2ab  +  b' 
'2ab  -4-  ¥ 


2a  -f  26  4-  c  2ac  +  26c  -f  c^ 

2aG  +  26(?  +  c^ 

16* 


186  EVOLUTION. 

Analysis.- -Proceeding  as  before,  we  obtain  two  terms  of  the  root, 
a  {-  b,  and  a  remainder  of  2ac  +  2bc  +  c-.  We  now  consider  a  -\-  b 
as  the  first  part  of  the  required  root,  and  write  2a  +  26,  or  twice  the 
pari  already  founds  for  a  divisor.  Dividing,  we  obtain  c,  the  next 
term  of  the  root,  which,  as  before,  we  phice  in  both  the  root  and 
divisor.  Multiplying  this  complete  divisor  by  c,  and  subtracting  the 
product  from  the  dividend,  we  have  no  remainder,  and  the  work  i» 
complete. 

18 T.  FroiL  these  illustrations  we  deduce  the  following 

Rule.  I.  Arrange  the  terms  according  to  the  powers  of 
some  letter,  beginning  with  the  highest,  and  write  the  square 
root  of  the  first  term  in  the  root, 

II.  Subtract  the  square  of  the  root  thus  found  from  the 
first  term,  and  bring  down  the  next  two  terms  for  a  dividend. 

III.  Divide  the  first  term  of  the  dividend  by  twice  the  root 
already  found,  and  write  the  result  both  in  the  root  and  in  the 
divisor, 

lY.  Multiply  the  divisor,  thus  completed,  by  the  term  of 
the  root  last  found,  and  subti^act  the  product  from  the 
dividend,  and  proceed  with  the  remainder,  if  any,  as  before. 

Note. — According  to  the  principles  established  in  (184),  every  square 
root  obtained  will  still  be  a  root,  if  all  the  signs  of  its  terms  be  changed. 


EXAMPLES   FOR   PRACTICE. 

1.  What  is  the  square  root  of  a*  +  4a^6  —  4a'^  4-  46^  — 
85  +  ^  ?  Ans,  a'  +  2b  —  2, 

2.  What  is  the  square  root  of  l—ib+4:b'-{-2y—4:by-i-7/? 

Ans.  1  —  2b  +  y. 

3.  Wh^i  is  the  square  root  of  4x*— 4x^  +  IS.r^—Gx  +  9  >' 

Ans,  2x^  —  x-i-3. 
4    What  is  the  square  root  of  4a:* — 16a;^-4-24a:^ — 16x-f4  ? 

Ans.  2x^  —  4a;  -f  2. 
5.  What  is  the  square  root  of  16a;*  +  24:x^  +  S9x^  -f  60a: 
4-  100  '/  Ans.  4x^  -{-  Sx  +  10. 


SQUARE  KOOT  OF  POLYNOMIALS.  187 

6    What  is  the  square  root  of  4x*  —  IG.c*  +  Sx'^  +  lOx  +  4  V 

Ans.  2x2  — 4x  — 2. 

7.  What  is  the  square  root  of  x'  +  2xy  +  y-  +  6x2  4-  %2 

8.  What  is  the  square  root  of  a^  —  ab  +  J 6'  ? 

9    What  is  the  square  root  of  r^  —  2  -f-  -  ? 

a       b  ha 

Ans.   , ,  or 7-. 

b       a         a       o 

2  11  2 

10.  What  is  the  square  root  of  x^  —  2x^2/    +  2/^  ^ 

^  i         i  i         i 

^ns.  x-^  —  t/-^,  or  1/*^  —  x-^. 

11.  What  is  the  square  root  of  1—4^  +  lOz'^— -202^  +  252:^ 
-.242^  +  l^z'  ?  Ans.  1  —  22-1-  82^  —  4z«. 

12.  W^hat  is  the  square  root  of  a«  —  6a^c  -f  15ttV  —  20aV 
4-  15aV  —  6ac^  +  c'  't  Ans,  a«  —  Sa^c  +  3ac^  —  d". 

13.  What  is  the  square  root  of  22-— 22  +  1-f  22/i-~2/i-f/i2? 

Ans,  2  +  h  —  1. 


SQUARE  ROOT   OF   NUMBERS. 

188,  In  extracting  the  square  root  of  numbers,  the  first 
thing  to  be  considered  is  the  relative  number  of  places  in  a 
given  number  and  its  square  root.  This  relation  is  exhibited 
in  the  following  illustrations  : 

Iloot«<.  Squares.  Roots.  Squares. 

Ill  1 

9  81  10  1,00 

99  98,01  100  1,00,00 

999  99,80,01  1000  1,00,00,00 

From  these  examples  we  perceive  that  a  number  consisting 
of  one  place  may  have  one  or  two  places  in  the  square ;  and 
that  in  all  cases  the  addition  of  one  place  to  the  root  adds  two 
places  to  the  square.     Hence, 


188  EVOLUTION. 

If  a  number  he  pointed  off  into  periods  of  two  figures 
each  J  commencing  at  the  right,  the  number  of  full  peinods, 
and  the  left  hand  full  or  jMrtial  period  will  be  equal  to  the 
number  of  places  in  the  square  root ;  the  highest  period 
answering  to  the  highest  figure  of  the  root, 

189*  The  square  of  any  numeral  quantity  may  be  formed 
after  the  manner  of  algebraic  squares. 

For  example,  let  a  ~  40,  and  6  =  7;  then  a  +  o  =  il. 
And  since  the  square  of  a  +  6  will  represent  the  square  of  47, 
W8  have 

a^  =  1600 
2a6=    560 
6^=      49 
a^  -f  2ab  +  6^  =  2200  =  (47)1 

Hence,  the  binomial  square  may  be  used  as  a  formula  for 
extracting  the  square  roots  of  numbers. 

1.  Extract  the  square  root  of  the  number  2209. 

OPERATION.  Analysis.  Here  are 

99  OQ  UO  -L  7  >-  17  ^^"^  P'^'^^^'  indicating 

^^,uy  i^u  -i~  ^  —  4-/  ^^^  places  in  the  root, 

Q?  =  1600  corresponding  to  tens 

2a.  =  80        609  ^"^  units.  The  great- 

2a  +  b  =  87      609  ^f  ^^";\''«i"  ^^  's  16, 

'  Its  root  IS  4,  or  4  tens 

=  40.  Hence,  a  =  40 
Then  2>a  =  80,  which  we  use  as  a  divisor  for  G09,  and  obtain  7  for  a 
quotient.  The  7  is  taken  as  the  value  of  6,  and  2a  -f  6,  the  com- 
plete divisor,  is  87,  which,  multiplied  by  7,  gives  the  last  two 
terms  of  the  binomial  square,  2a5 -{- 5^  =  609,  and  the  entire  root, 
40  4-  7  =  47,  is  found. 

Arithmetically,  a  may  be  taken  as  4  instead  cf  40,  and  we 
may  write  16  in  hundreds'  place,  instead  of  1600,  the  ciphers 
being  superfluous.  Then  2a  will  be  8  instead  of  80,  and  in 
dividing,  we  say  8  is  contained  in  60  (not  in  609)  7  times. 

If  the  given  number  consists  of  more  than  two  periods,  we 
obtain  the  two  superior  figures  of  the  root  from  the  first  two 


SQUARE  ROOT  OF  NUxMBERS.  1S9 

periods,  as  before,  and  bring  down  another  period  to  the 
remainder.  We  then  consider  the  root  already  found  as  one 
qaantity,  and  treat  it  as  one  figure. 

2.  What  is  the  square  root  of  390424  ? 

OPERATION.  Analysts.      Disregarding    the   local 

'^o  G  <  cA^por,  value   of  the  fl,i!;ures,  we    have    a  =  G, 

;iJ  J-t,^4:(po:^  2fl  =  12,  and  12  in  39,  3  times,  which 

36 "  gives  5  =  3.    We  next  suppose  a  =  63, 

123      "S94  ^"^  -^  —  ^^^  5  ^"^  1^^  ^^  ^^-'  2  *^^^'^» 

o/»rk  or  the  second  value  of  5  =  2.     In  the 

same    manner,    we   would    repeat   the 

1262        25  24-  formuI?«  of  a  binomial  square  as  many 

25  24  times  as  we  have  periods.    It  is  evident 

ihat  we  may  obtain  the  divisor  126  from 
the  last  complete  divisor  123  simply  by  doubhng  its  last  figure  3 ; 
and  thus  the  divisors  may  be  derived  each  from  the  next  preceding, 
successively. 

From  these    examples    and    Illustrations   we   deduce    the 

following- 

Rule.  I.  Point  the  given  nu7nher  off  into  periods  of  tivo 
figures  each,  counting  from  the  units^  place  to  the  left  and 
right. 

II.  Find  the  greatest  2^67 feet  square  in  the  left-hand  pe- 
riod, and  write  its  root  for  the  first  figure  in  the  required 
root ;  subtract  the  square  of  this  figure  from  the  first  joeriod^ 
and  to  the  remainder  bring  down  the  next  period  for  a 
dividend. 

III.  Double  the  i^oot  already  found,  and  write  the  result 
on  the  left  for  a  divisor  ;  find  how  many  times  this  divisor  is 
contained  in  the  dividend,  exclusive  of  the  right-hand  figure, 
and  place  the  result  in  the  root  and  at  the  right  of  the  divisor. 

lY.  Multiply  the  divisor  thus  completed  by  the  last  figure 
of  the  root ;  subty^act  the  product  from  the  dividend,  and  to 
the  remainder  bring  down  the  next  p)eriod  for  a  new  divi- 
dend. 

Y.  Double  the  right-hand  figure  of  the  last  complete  divi- 
sor for  a  new  divisor,  and  continue  the  operation  as  before. 


190  EVOLUTION.  r 

EXAMPLES   FOR   PRACTICE. 

3.  What  is  the  square  root  of  883G  ?  Ans.  94. 

4.  What  is  the  square  root  of  106929  ?  Ans.  327. 

5.  What  is  the  square  root  of  47829G9  ?  Ans.  2187. 
6  What  is  the  square  root  of  43046721  ?  Ans.  6561. 
:    What  is  the  square  root  of  387420489  ? 

Ans.  19683. 

8.  What  is  the  square  root  of  1209996225  ? 

Ans.  34785. 

9.  What  is  the  square  root  of  6596038656  ? 

Ans.  81216. 

10.  What  is  the  square  root  of  342694144  ? 

Ans.  18512. 

U.  What  is  the  square  root  of.  2573733560796  ? 

.4ns.  1604286. 

12.  What  is  the  square  root  of  10.4976  ?         Ans.  3.24. 

13.  What  is  the  square  root  of  3271.4207  ? 

Ans.  57.19  -f . 

14.  What  is  the  square  root  of  4795.25731  ? 

Ans.  69.247+. 

15.  What  is  the  square  root  of  .0036  ?  Ans.  .06. 

16.  What  is  the  square  root  of  .00032754  ? 

Ans.  .01809  +. 

17.  What  is  the  square  root  of  .00103041  ? 

Ans.  .0321. 

Note. — If  both  terms  of  a  fraction  are  perfect  squares,  or  if  the  frao- 
tion  can  be  reduced  to  terms  which  are  squares,  the  root  may  be  ob- 
tained b}^  the  rule  for  algebraic  fractions.  Otherwise,  the  fraction  may 
be  reduced  to  a  decimal. 

18.  What  is  the  square  root  of  §f  ?  Ans.  |. 

19.  What  is  the  square  root  of  ^^^j  ?  Ans.  ■^-^,     j 

20.  What  is  the  square  root  of  ^^i£^  ?  Ans.   ^%\.    ^ 


SQUARE  ROOT  OF  NUMBERS  191 

21.  What  is  the  square  root  of  yVg  ? 

Observe  -j^^^  =  ||.       Hence,  the  square  root  is  |, 

22.  What  is  the  square  root  of  ff §1  ?  Ans.  -§. 

23.  What  is  the  square  root  of  f|gi  ?  Ans,  f. 
24  What  is  the  square  root  of  |  '/  Ans.  866  +. 
25.   What  is  the  square  root  of  ^  ?             ^ns^ .8819  H-. 

CUBE  ROOT  OF  POLYNOMIALS.*  ^ 

lf^<0.  We  may  derive  the  method  of  extracting  the  cube 
root  of  an  algebraic  quantity  in  a  manner  similar  to  that 
pursued  in  square  root,  by  analyzing  and  retracing  the  combi- 
nation of  terms  in  the  binomial  cube.  Forming  the  cube  of 
a  -\-  b,  we  have 

(a  +  by  =a'  +  2>a'b  +  Sab'  +  5^ ; 
from  which  we  see  that 

I.  The  first  term  of  the  power  is  the  cube  of  the  first  term 
of  the  root;  and 

II.  The  second  term  of  the  poiveris  three  times  the  square 
9f  the  first  term  of  the  root  multiplied  by  the  second. 

It  is  evident,  therefore,  that  to  find  the  first  term  of  the  root, 
we  must  extract  the  cube  root  of  the  first  term  of  the  power ; 
and  to  find  the  second  term  of  the  root,  we  must  divide  the 
second  term  of  the  power,  3a^&,  by  three  times  the  square  of 
the  first  term  of  the  root,  8a'^ ;  thus, 

3a^&  --  8a-  =  b. 

The  last  three  terms  of  the  power  may  be  factored  as 
follows : 

(Sa^  +  3a6  +  b'')b. 

To  reproduce  these  terms  from  the  divisor  already  found  and 
the  root,  we  must  complete  our  partial  divisor,  3a^,  by  the  addi- 

*  We  are  Indebted  to  J.  C.  Porter,  A.M.,  of  the  Clinton  Liberal  Institute,  for  the 
valuable  method  of  Cube  Root  presented  here  and  in  the  Practical  Arithrootia.  It  is 
an  extension  of,  and  improvement  upon  Horner's  Method,  and  secures  the  result  with 
less  labor  than  any  other  method  heretofore  presouted. 


192  EVOLUTION. 

tion  of  Sab  +  b^,  and  multiply  the  divisor  thus  completed  by  b. 
Putting  the  correction,  oab  +  b"^,  under  the  form  of  (3a  +  b)b, 
we  shall  have, 

3a^  =  Trial  divisor. 
3a    -{  b  =  First  factor  of  correction. 
oab  4-  6^  =  Correction  of  trial  divisor. 
3a^  +  Sab  +  b^  —  Complete  divisor. 

1.  .Find  the  cube  root  of  a^  +  Sa'b  +  Sab^  +  b\ 

OPERATION. 

a'  +  Sa'b  +  Sab'  +  b'  {a-hb 

^ 

|3a^  Sa:'b  +  Sab'  -f  b' 

da  +  b     Sab  +  b'  \sa'  +  3a6  +  ¥      Sa'b  +  3a6^  +  b' 


Analysis.  Taking  the  cube  root  of  a**,  we  obtain  a,  the  first  terra 
of  the  root.  Subtracting  the  cube  of  a  from  the  given  polynomial, 
we  have  Za^b  -j-  3a6^  -f  b^  for  a  remainder  or  dividend.  We  next  write 
Za'^  at  the  left  of  the  dividend  for  a  trial  divisor.  Dividing  the  first 
term  of  the  dividend,  we  obtain  b,  the  second  term  of  the  root.  We 
next  multiply  the  former  term  of  the  root  by  3,  and  annex  the  latter 
term,  6,  and  obtain  3a  -\-  b,  the  first  factor  of  the  correction  to  the 
trial  divisor.  Multiplying  this  by  b,  we  have  oab  +  b'\  the  correction. 
Adding  this  to  the  trial  divisor,  we  have  3a^+  3a6+  6^  the  complete 
divisor.  Multiplying  the  complete  divisor  by  the  last  term  of  the 
root,  and  subtracting  the  result  from  the  dividend,  we  have  no 
remainder,  and  the  work  is  complete. 

Again,  let  us  form  the  cube  of  any  polynomial,  as  a  4-  6  +  c, 
in  the  following  manner  : 

Assume  s  =  a  +  b,  the  first  part ; 
c  =  the  second  part. 

Then  (s  +  c^  =  s^  +  Ss'c  +  Ssc'  +  c\ 

The  first  two  terms  of  the  root,  regarded  as  one  part,  sus- 
tain the  same  relation  to  the  third,  as  the  first  sustains  to  the 
second :  and  so  on. 

The  binomial  cube,  therefore,  furnishes  the  method  of  ex- 
tracting any  cube  root  whatever,  by  treating  the  root  already 
found,  at  each  step,  as  a  simple  term. 


CUBE  ROOT  OF  POLYNOMIALS.  193 

2.  What  if5  the  cube  root  of  ^  —  40^^^  +  6a^  +  96a;  —  64  ? 


OPERATION. 


Sa;«+2aj  6ar»+42;g 


33;Q+6a^-4  — 12a;«— 24a;+16 


3a;*  +  6a;»+  4a;«  6a;»+12a;*+  8a^ 


3x*+12»»+12a;3  _i2a;<— 48a:'+96a5— 64 

3a;<+12a;«— 24a;  +16        —12ai*—4:Sx^-\-96x—64. 


Analysis.  Since  it  was  shown,  in  involution,  that  the  exponents 
of  any  letter  in  a  power  form  a  regular  series,  we  arransje  the  terms 
according  to  the  powers  of  x.  The  cube  root  of  a^  is  aj^  the  first  term 
of  the  root;  subtracting  the  cube  of  x^  from  the  polynomial,  and 
arranging  the  remainder  according  to  the  powers  of  x,  we  have  6x^  — 
40.r*  +  96a;  —  64  for  a  dividend.  We  next  write  3  times  the  square 
of  a;^  or  3a:*,  for  a  trial  divisor  ;  and  dividing  6x*,  the  first  term  of  the 
dividend,  we  obtain  2x  for  the  second  term  of  the  root.  Having  found 
the  second  term,  we  must  complete  our  divisor  as  in  the  first  exam- 
ple. Therefore,  to  3  times  the  first  term  we  annex  the  second,  and 
obtain  Zx"^  +  2x,  the  first  factor;  and  multiplying  this  by  the  second 
term,  we  have  6ar*  -f  4:x^  for  the  correction  to  the  trial  divisor.  Add- 
ing, we  have  Sa:*  -j-  6a:^  -f  ^^'^i  the  complete  divisor.  Multiplying 
this  by  the  second  term,  2x,  and  subtracting  the  product  from  the 
dividend,  we  have  for  a  new  dividend,  — 12a:*  —  48a:^  +  96a;  —  64. 

Now,  since  the  two  terms  of  the  root  already  found,  considered  as 
one  part,  sustain  the  same  relation  to  the  third  term,  as  the  first 
term  sustains  to  the  second,  the  trial  divisor  to  obtain  the  third  term 
will  be  3  times  the  square  of  the  first  two  terms,  or  3(a;2  -{-  2xy  = 
Z.'c^  -{-  12a:^  -f  12a:2«  This  quantity  is  found  in  the  operation  by  add- 
ing together  4ic*,  the  square  of  the  last  term  of  the  root;  6x^  -f-  4a:^ 
the  correction  ;  and  3a;*  -f  6a:^  -f-  4a;2,  the  first  complete  divisor. 
Dividing  —  12a;*,  the  first  term  of  the  dividend,  by  3a;*,  the  first  term 
of  the  divisor,  we  obtain  — 4,  for  the  third  term  of  the  root. 

To  find  a  correction  of  the  trial  divisor,  the  first  factor  will  be  the 
last  term,  — 4,  annexed  to  three  times  the  former  terms  of  the  root, 
or  3a;^  -f  6a;  —  4.  This  quantity  is  found  in  the  operation  by  taking 
the  first  factor  of  the  last  correction,  with  its  last  term  multiplied  by 
3,  and  annexing  the  — 4.  Multiplying  this  by  — 4,  we  obtaia 
—  12a;2  —  24a;  +  16,  for  the  correction.  Adding  this  to  the  trial 
divisor,  we  have  3a;*  +  12a;^  —  24a;  +  16,  for  the  complete  divisor. 
Multiplying  this  by  — 4,  and  subtracting  the  result  from  the  divi- 
dend, we  have  no  remainder,  and  the  work  is  complete. 
17  N 


194  EVOLUTION. 

191.  From  these  illustrations  we  deduce  the  foLowing 

KULE.  I.  Arrange  the  polynomial  according  to  the 
powers  of  some  letter^  and  write  the  cube  root  of  the  first 
term  in  the  root, 

II.  Subtract  the  cube  of  the  root  thus  found  from  the  poly^ 
nomial,  and  arrange  the  remainder  for  a  dividend. 

III.  At  the  left  of  the  dividend  write  three  times  the  square 
of  the  root  already  found  for  a  tr^ial  divisor ;  divide  the  first 
term  of  the  dividend  by  this  divisor,  and  write  the  quotient 
for  the  next  term  of  the  root, 

lY.  To  three  times  the  first  term  of  the  root  annex  the  last 
term,  and  write  the  result  at  the  left,  and,  one  line  below,  the 
trial  divisor  ;  multiply  this  binomial  factor  by  the  last  term 
of  the  root,  for  a  correction  to  the  trial  divisor  ;  add  the  cor- 
rection, and  the  result  will  be  the  complete  divisor. 

Y.  Multiply  the  complete  divisor  by  the  last  term  of  the 
"^oot,  subtract  the  product  from  the  dividend,  and  arrange  the 
remainder  for  a  new  dividend. 

YI.  Add  together  the  square  of  the  last  term  of  the  ro<?t, 
the  last  correction,  and  the  last  complete  divisor,  for  a  new 
trial  divisor,  and  by  division  obtain  another  term  of  the 
root. 

YII.  Take  the  first  factor  of  the  last  correction  with  its  last 
term  multiplied  by  3,  ajid  annex  to  it  the  last  term  of  the  root^ 
for  the  first  factor  of  the  correction  to  the  neio  trial  divisor, 
with  which  proceed  as  in  the  former  steps,  till  the  work  is 
completed. 

Note. — The  first  term  of  the  remainder,  when  properly  arranged,  will 
"be  that  term  which  contains  the  highest  power  of  the  leading  letter  of  the 
root,  or  of  the  arranged  polynomial. 

EXAMPLES   FOR   PRACTICE. 

1.  What  is  the  cube  root  of  8  -f  12a  +  6a^  +  «^  ? 

Ans.  2  +  «. 

2.  What  is  the  cube  root  of  27a^  +  108a'  +  144a  +  64  ? 

Ans.  &a  'f  4. 


CUBE  ROOT  OF  NUMBEUS.  ^95 

8.  What  is  the  cube  root  of  a^  —  Wx  +  12ax2  —  8;c3? 

An8,  a  —  2x 

4.  What  is  the  cube  root  of  ic«  —  3x«  +  5a;»  --  3x  —  1  ? 

Ans,  x^  —  X  —  1. 

5.  What  is  the  cube  root  of  a^  —  Qa'b  +  12ah''  —  86'  ? 

Ans,  a  —  26. 

3        1 

6.  What  is  the  cube  root  of  cr^  +  3a:  +  -  +  —  ? 

X         x^ 

Ans.  X  -I — . 

X 

7.  What  is  the  cube  root  of  x^  +  ox^  +  6x^  +  10x«+  12x5  + 
12x*  +  lOx^  +  6x2  +  3x  +  1  ? 

Ans.  x^  +  x'^  +  X  +  1. 

8.  What  is  the  cube  root  of  a»  —  Sa^  4-  8a«  —  Qa^  —  6a*  -|- 
8a'  — .  8a  +  1  ? 

^ns.  a'  —  a'^  —  a  4-  1. 

CUBE  ROOT  OF  NUMBERS. 

19S.  To  apply  the  binomial  cube  as  a  formula  for  the  ex- 
traction of  the  cube  root  of  numbers,  we  must  first  ascertain 
the  relative  number  of  places  in  a  cube  and  its  root.  This  re- 
lation will  be  seen  in  the  following  examples. 


Boots. 

Cubes. 

Roots. 

Cubes. 

1 

1 

1 

1 

9 

729 

10 

1,000 

99 

970,299 

100 

1,000,000 

999 

997,002,999 

1000 

1,000,000,000 

From  these  illustrations,  we  perceive  that  a  number  con- 
sisting of  one  place,  may  have  from  one  to  three  places  in  its 
cube ;  and  that  in  all  cases  the  addition  of  one  place  to  the 
root  adds  three  places  to  the  cube.     Hence, 

If  a  number  he  pointed  off  into  three-figure  periods,  com- 
mencing  at  the  right,  the  number  of  full  periods,  and  the 


196  EVOLUTION. 

left-hand  full  or  partial  period,  will  indicate  the  number  of 
places  in  its  cube  root ;  the  highest  period  answering  to  the 
highest  figure  of  the  root, 

lf>3.  To  form  the  cube  of  a  number,  let  a  =  50,  and  b  =  4. 
Then  a  +  6  =  54  ;  and  cubing,  we  have, 

a«=  125000 
3a^6=  30000 
8a&2  ==      2400 

___J h^=-  64 

(a  +  6)8  ==  a»  +  Za'^h  +  Sab^  +  6^  =  157464  =  (54)^ 

Hence,  in  the  cube  of  a  number. 

The  figures  of  the  root,  with  their  local  values,  have  the 
same  combinations  as  the  terms  of  an  algebraic  quantity. 

1    What  is  the  cube  root  of  157464  ? 

OPERATION.  Analysis.      Pointing   off  the 

157  464  I  54  number,  the  two   periods   show 

-jo;;  that   there  will  bo  two  figures, 

' — oT/<7>l *^^^  ^^^^  units,  in  the  root.  Sincn 

7500     3l464  ^^le  highest  figure  of  the  root  cor- 

154     616    8116     32464  responds  to  the  highest  period  of 

the  power,  we  find  the  greatest 
perfect  cube  in  the  first  or  left  hand  period,  which  is  125,  and  place 
5,  its  root,  for  the  figure  of  the  required  root.  Subtracting  the  cube 
number  125  from  the  first  period,  and  bringing  down  the  next  period, 
we  have  32464  for  a  remainder  or  dividend.  Since  the  figures  in  a 
cube  root,  with  their  local  values,  have  the  same  combinations  in  the 
cube  as  the  terms  of  an  algebraic  quantity,  we  write  at  the  left  of 
the  dividend  three  times  the  square  of  the  root  already  found,  or  75, 
with  tvro  ciphers  annexed,  for  a  trial  divisor.  Dividing,  we  obtain  4, 
Tor  the  next  figure  of  the  root.  To  complete  the  divisor,  we  multiply 
the  first  figure  of  the  root  by  3,  and  annex  the  last,  and  obtain  154  for 
the  first  factor  of  the  correction.  Multiplying  this  number  by  4,  wo 
have  616,  the  correction  to  the  trial  divisor.  Adding,  we  have  8166, 
the  complete  divisor.  And  multiplying  this  by  4,  and  subtracting 
tb.e  product  from  the  dividend,  there  is  no  remainder,  and  the  work 
is  complete. 


CUBE  ROOT  OP  NUMBERS. 


197 


OPERATION. 

12,812,904|234 

8 

63 

189 

1200   4812 
1389   4167 

694 

2776 

158700  645904 
161476  64e5904 

2.  Wliat  is  the  cube  root  of  12812904  ? 

Analysis.  The  great- 
est cube  in  the  first  pe- 
riod is  8,  and  its  root  is  2, 
■which  we  write  for  the 
first  figure  of  the  required 
root.  Subtracting  8,  and 
bringing  down  the  next 
period,  we  have  4812  for 
a  dividend.  Annexing 
two  ciphers  to  3  times 
the  square  of  2,  we  have 
1200  for  a  trial  divisor.  Dividing, we  obtain  3,  the  next  figure  of  tho 
root.  To  complete  the  divisor,  we  have,  bj  the  same  method  as  be- 
fore, 63  for  the  first  factor  of  the  correction,  and  189  for  the  correc- 
tion. Adding  the  correction,  we  obtain  1389  for  the  complete  divisor. 
Multiplying  this  by  3,  subtracting  the  product  from  the  dividend, 
and  bringing  down  the  next  period,  we  have  645904  for  a  new  divi- 
dend. As  in  the  algebraic  method,  we  add  9,  the  square  of  the  last 
root  figure;  189,  the  last  correction;  and  1389,  the  last  complete  divi- 
sor ;  and  annex  two  ciphers,  for  a  new  trial  divisor.  Dividing,  we 
obtain  4,  the  next  figure  of  the  root.  AVe  then  take  the  first  factor 
of  the  last  correction,  with  its  last  figure  multiplied  by  3,  and  annex 
the  last  root  figure,  4,  and  obtain  694  for  the  first  factor  of  the  new 
correction.  Multiplying  this  by  4,  we  have  2776,  the  correction. 
Then  completing  the  divisor,  multiplying  by  the  last  root  figure,  and 
subtracting  the  product  from  the  dividend,  we  have  no  remainder, 
and  the  work  is  complete. 


From  these  examples  we  derive  the  following 

Ktjle.  I.  Point  off  the  given  number  into  periods  of 
three  figures  each,  counting  from  units''  place  to  the  left  and 
right. 

II.  Find  the  greatest  cube  in  the  left-hand,  period,  and 
place  its  root  for  the  first  figure  of  the  required  root. 
Subtract  the  cube  from  the  first  period,  and  to  the  remainder 
bring  doivn  the  next  period  for  a  dividend, 

III.  At  the  left  of  the  dividend,  write  three  times  the 
square  of  the  root  already  found,  and  annex  two  ciphers  for 

17* 


198  EVOLUTION. 

a  Mai  divisor;  divide  the  dividend ^  and  write  the  quotient 
for  the  next  term  of  the  root. 

lY.  To  three  times  the  first  figure  of  the  root  annecb  the 
lastf  and  place  the  result  at  the  left,  and  one  live  below  the 
trial  divisor  ;  multiply  the  factor  by  the  last  7^oot  figui^e,  for 
a  correction  to  the  tr^ial  divisor  ;  add  the  correction^  and  the 
result  will  be  the  complete  divisor. 

Y.  Multiply  the  complete  divisor  by  the  last  figure  of  the 
root,  subtract  the  product  from  the  dividend,  and  to  the 
remainder  bring  down  another  period  for  a  new  dividend. 

YI.  Add  together  the  square  of  the  last  figure  of  the  root, 
the  last  correction,  and  the  last  comp)lete  divisor,  and  annex 
two  ciphers  for  a  new  trial  divisor  ;  and  by  division  obtain 
another  figure  of  the  root. 

YII.  Take  the  first  factor  of  the  last  correction,  with  its 
unit  figure  multiplied  by  3,  and  annex  the  last  figure  of  the 
root,  for  the  first  factor^  of  the  correction  to  the  new  trial 
divisor ,  with  which  proceed  as  in  the  former  steps  till  the 
work  is  complete. 

Notes.  1.  If  at  any  time  the  product  is  greater  than  the  dividend, 
diminish  the  corresponding  root  figure  and  correct  the  erroneous  work. 

2.  If  a  cipher  occur  in  the  root,  annex  two  more  ciphers  to  the  trial 
divisor,  bring  down  another  period  in  the  dividend,  and  proceed  as  before. 

EXAMPLES  FOR  PRACTICE. 

3.  What  is  the  cube  root  of  148877  ?  Ans.  53, 

4.  What  is  the  cube  root  of  571787  ?  Ans.  83. 

5.  What  is  the  cube  root  of  1367631  ?  Ans.  111. 

6.  What  is  the  cube  root  of  2048383  ?  Ans.  127. 

7.  What  is  the  cube  root  of  16581375  ?  Ans.  255. 

8.  What  is  the  cube  root  of  44361864  ?  Ans.  354. 
9  What  is  the  cube  root  of  100544625  ?  Ans.  465 

10.  What  is  the  cube  root  of  12358-435328  ?  Ans.  2312 


CUBE  ROOT  OF  NUMBERS.  199 

11.  What  is  the  cube  root  of  999700029999  ?  Ans.  9999. 

12.  What  is  the  cube  root  of  2456  ?         Ans,  13.491+. 

13.  What  is  the  cube  root  of  .004019679?       Ans.  .159. 

14.  What  is  the  cube  root  of  2287.148  ?  Ans.  13.175  +. 

CONTRACTED    METHOD. 

104.  The  methods  of  direct  extraction  of  the  cube  root 
of  surd  numbers  are  all  too  tedious  to  be  much  used,  and  seve- 
ral eminent  mathematicians  have  given  more  brief  and  practi- 
cal methods  of  approximation. 

Olc  of  the  most  useful  methods  may  be  investigated  as 
follows : 

Suppose  a  and  a  +  c  two  cube  roots,  c  being  very  small  in 
relation  to  a ;  a*  and  a^  +  Sa'-c  +  Sac^  +  c^  are  the  cubes  of  the 
supposed  roots. 

Now,  if  we  double  the  first  cube  (a^),  and  add  it  to  the 
second,  we  shall  have 

8a^  -f  Sa'-c  -f  3ac'  +  c*. 

If  we  double  the  second  cube  and  add  it  to  the  first,  we 

shall  have 

Sa'  +  6a'c  +  6ac^  +  2c\ 

As  c  is  a  very  small  fraction  compared  to  a,  the  terms  con- 
taining c''  and  c^  are  very  small  in  relation  to  the  others ;  and 
the  relation  of  these  two  sums  will  not  be  materially  changed 
by  rejecting  those  terms  containing  c^  and  c^,  and  the  sums 

will  then  be 

And  oa^  +  Qd^c, 

The  ratio  of  these  terms  is  the  same  as  thg  ratio  of  a  +  oto 
a  f  2c. 

c 


Or  the  ratio  is  1  + 


a  +  c 


But  the  ratio  of  the  roots  a  to  a  +  c,  is  1  H 


200  EVOLUTION. 

Observing  again,  that  c  is  supposed  to  be  very  small  in  rela- 

c  c 

tion  to  a,  the  fractional  parts  of  the  ratios and     are 

'  ^  a  -\-  G  a 

both  small,  and  very  near  in  value  t<)  each  other.  Hence,  we 
have  found  an  operation  on  two  cubes  which  are  near  each 
other  in  magnitude,  and  that  will  give  results  very  near  m 
proportion  to  their  roots ;  and  by  knowing  the  root  of  one  of 
the  cubes,  by  this  ratio  we  can  find  the  other.  And  as  this 
relation  will  still  exist  if  one  of  the  roots  is  a  surd,  the  propor- 
tion will  furnish  a  method  of  approximating  to  values  of  surds. 
For  example,  let  it  be  required  to  find  the  cube  root  of  28, 
true  to  4  or  5  places  of  decimals.  Since  27  is  a  cube  near  in 
value  to  28,  the  root  of  which  we  know  to  be  3, 

Assume  a^  =  27,         or         a  =  3 . 

(a  +  cf  =  28,         or  a  +  c  =  ^28- 


Then 

27 

28 

2 

2 

54 

56 

Add 

28 

27 

Sums 

82 

:    83 

3  :  a  -I-  <?  very  nearly. 

Or,  (a  +  c)  =  y/  ^  3.03658  +,  which  is  the  cube  root 
of  28,  true  to  6  places  of  decimals. 

By  the  laws  of  proportion,  which  we  hope  more  fully  to  in- 
vestigate in  a  subsequent  part  of  this  work,  the  above  pro- 
portion, 82  :  83  : :  a  :  a  -f-  c, 
may  take  this  form,          82  ;     1  : :  a  :  c,         c  being  a  cor- 
rection to  the  known  root,  a.  ' 

Hence        c  =  ^^^  =  .03658  +  ; 
And  a  -\-  c  =  3.03658  +,  as  before. 

195.  From  this  investigation,  we  deduce  the  following  rule 
for  finding  approximate  cube  roots : 

Ktjle.      Take  the  nearest  rational  cube  to  the  given  nam- 
bery  or  assume  a  root  and  cube  it.     Double  this  cube,  ana 


CUBE  ROOT  OF  NUMBERS.  201 

add  the  number  to  it ;  also  double  the  number  and  add  the 
assumed  cube  to  it.  Then,  by  proportion,  the  first  sum  is  to 
the  secondj  as  the  known  root  is  to  the  required  root. 

Or,  The  first  sum  is  to  the  difference  of  the  two  sums,  an 
the  known  root  is  to  a  coi^rection  to  the  known  root, 

EXAMPLES   FOR  PRACTICE. 

1.  What  is  the  approximate  cube  root  of  122  ? 

Ans,  4.95967  +. 

2.  What  is  the  cube  root  of  10  ?  Ans.  2.65441  +. 
Note. — Assume  2.1  for  the  root,  then  9.261  is  its  cube. 

3.  What  is  the  approximate  cube  root  of  720  ? 

Ans.  8.9628  +. 

4.  What  is  the  approximate  cube  root  of  845  ? 

Ans.  7.01357  +. 

5.  What  is  the  approximate  cube  root  of  520  ? 

Ans.  8.04145+. 

6.  What  is  the  approximate  cube  root  of  65  ? 

Ans.  4.0207  +. 

7.  What  is  the  approximate  cube  root  of  16  ? 

The  cube  root  of  8  is  2,  and  of  27  is  3  ;  therefore  the  cube 
root  of  16  is  between  2  and  3.  Suppose  it  2.5.  The  cube  of 
this  root  is  15.625,  which  shows  that  the  cube  root  of  16  is  a 
little  more  than  2.5,  and  by  the  rule 


15.625 

2 

16 

2 

31.250 
16 

32 
15.625 

47.25 

47.625 

47.25 

47.25 

:         375  : :  2.5  :  .01984 

Assumed  root, 
Correction, 

2.50000 
.01984 

Approximate  root,  2\  51984 


2:02  EVOLUTION. 

We  give  the  last  as  an  example  to  be  followed  in  most 
cases  where  the  root  is  about  midway  between  two  integral 
numbers. 

This  method  may  be  used  with  advantage  to  extract  tlie 
root  of  perfect  cubes,  when  very  large,  as  will  be  seen  in  the 
examples  which  follow. 

8.  The  number  22.069.810.125  is  a  cube  ;  required  its  root 

Analysis.  Dividing  this  cube  into  periods,  we  find  that  the  root 
must  contain  4  figures,  the  superior  period  is  22,  the  cube  root  of  22 
is  near  3,  and  of  course  the  whole  root  near  3000 ;  but  it  is  less  than 
3000.  Suppose  it  2800,  and  cube  this  number.  The  cube  is 
21952000000,  which,  being  less  than  the  given  number,  shows  that 
our  assumed  root  is  not  large  enough. 

To  apply  the  rule,  it  will  be  sufficient  to  take  six  superior  figures 
of  the  given  and  assumed  cubes.     Then  by  the  rule, 

219520    220698 


439040 

441396 

220698 

219520 

659738 

660916 

659738 

659738 

:   1178  : :  2800 

2800 

942400 

2356 

659 

738)3298400(5 
3298690 

Assumed  root,  2800 

Correction,      5 

True  root,    2805 

The  result  of  the  last  proportion  is  not  exactly  5,  as  will  be  seen 
by  inspecting  the  work  ;  the  slight  imperfection  arises  from  the  rule 
being  approximate,  not  perfect. 


REDUCTION  OF  RADICALS.  OQS 

TvfoTE.  — When  we  have  cubes,  "we  can  always  decide  ilie  unit  ngure 
by  inspection,  and,  in  the  last  example,  the  unit  tigure  in  the  cube  beincr 
5,  the  unit  ligure  in  the  root  must  be  5,  as  no  other  figure  when  cubed 
will  give  5  in  the  place  of  units. 

9.  The  number  41135081408  is  a  perfect  cube ;  required 
its  root.  A71S.  3452. 

10.  The  number  125525735343  is  a  perfect  cube  ;  required 
its  rooL  Ans,  5007. 

REDUCTION  OF  RADICALS. 

196.  A  Radical  Quantity  is  a  root  merely  indicated  by  the 
radical  sign  or  by  a  fractional  exponent;  as  2%/^,  d^ a^ — 26, 

The  quantity  or  factor  placed  before  a  radical  is  its  coeffi- 
cient. Thus,  2,  5,  and  c,  in  the  above  examples,  are  the  coeffi- 
cients of  the  radicals 

197.  The  degy-ee  of  a  radical  quantity  is  denoted  by  the 
radical  index,  or  by  the  denominator  of  the  fractional  expo- 
nent.    Thus, 

/ —     1  1 

V  a6,  x'^f  (a  +  b)^  are  radicals  of  the  2d  degree; 

"v^m,  ^ a^y  (2xy  are  radicals  of  the  3d  degree; 

Vx\  y^f  (a  —  c)n  are  radicals  of  the  nth  degree. 

198.  Similar  Radicals  are  those  having  the  same  quantity 
under  a  radical  sign  of  the  same  index.  Thus,  3\^a^ — b, 
—  v^a2  —  ^^  j^Q^j  5(^^2  —  i^y^  ^j.Q  similar  radicals. 

199.  Reduction  of  Radical  Quantities  is  the  process  of 
changing  their  forms  without  altering  their  values. 

CASE   I. 

300.  To  reduce  a  radical  to  its  simplest  form. 

A  radical  is  in  its  simplest  form  when  it  contains  no  perfect 
powers  corresponding  to  the  degree  of  the  radical      It  has 


\ 


204  RADICAL  QUANTITIES. 

been  shown  (185)  that  the  root  of  a  quantity  is  obtained  by 
dividing  the  exponent  of  each  factor  by  the  index  of  the  re- 
quired root ;  by  which  process  the  root  of  each  factor  is  taken 
separately.     Hence, 

The  root  of  a  quantity  is  equal  to  the  product  of  the  roots 
of  its  component  factors. 


1    E  educe  \^l^a^x  to  its  simplest  form. 

OPERATION.  .  ^.  ^,  ,..    , 

Analysis,     omce  the  radical 

y^TSa^=  \^%hd'  X  3^  is  of  the  second  degree,  we  se- 

^9-2  w  \/qJ        parate  the  factors  of  the  quan- 

X—  tity  under  the  radical  sign  into 

^^    ^      '  two  groups,  one  of  which,  25a'2, 

contains  2XS.Sh^ perfect  squares,  and  the  other,  3.^,  all  the  surds  in  the 

quantity.     And  since  the  root  of  a  quantity  is  equal  to  the  product 

of  the  roots  of  its  component  factors,  we  extract  the  square  root  of 

the  rational  part,  25<2^  and  obtain  5a,  and  multiplying  this  result  by 

the  indicated  root  of  the  other  part,  wo  obtain  5a\/3a;,  the  simplest 

form  of  the  radical. 

2.  Reduce  S^^a""  —  a^h  to  its  simplest  form. 

OPERATION.  Analysis.      Since  the 

^  3. ^  3 radical  is  of  the  third  de- 

V  a         ab  =  OVa(a  —  b)  gree,    we    separate    the 

=  5'v/a^  X    v^g        h  quantity  under  the  sign 

3. into  two  factors,  one  of 

=  5aVa  — 6  y^lxioh,   a\    is  a  perfect 

cube.     Taking  the  cube 

root  of  this  factor,  and  multiplying  this  root,  a,  the  coefficient,  5, 

and  the  surd,  >/ a  —  6,  together,  we  have  Sav^a  —  6,  the  simplest 

form  of  the  radical. 

From  these  illustrations  we  deduce  the  following 

R.XJLE.  I.  Separate  the  factors  of  the  quantity  under  the 
radical  sign  into  two  groups^  one  of  which  shall  contain  all 
the  perfect  powers  corresponding  in  degree  ivith  the  radical. 

II.  Extract  the  root  of  the  rational  part,  and  multiply  the 
root,  ooejficientj  and  surd  or  radical  part  together. 


REDUCTION. 


205 


EXAMPLES   FOR   PRACTICE. 

Reduce  the  following  radicals  to  their  simplest  form , 

Ans.  a\^hc. 


8.  v^^^ 

4.  2v^a^ 

5.  3V50^ 

6.  ai/TO^. 


Ans.  2xy\/ X, 


Ans, 


s/W. 


7.  6v/81m\ 


8.   '^d'—a'i 


9.  xy^  x^if-  —  a^y. 
10.  Wriah'&, 


11.  2av/l47aV?/. 

12.  5^125x. 


13.  2cv/82^\ 

14.  (a  +  6)  s/d?  —  2a?h  +  ah\ 

15.  {a  —  h)s/a'h  +  2ah^'{-h\ 

16.  cZ  ''/^^t/  —  2a;y  +  a;?/'. 

17.  (ISOa;^!/)* 

18.  (24^2/^2)^ 

19.  (54am«)*. 


20.  (aV  — a^62') 


i 


Ans.  av^l  —  c?. 

<4ns.  ^y  V  X  —  2/. 

u4ns.  24&cv^2ac. 

^ns.  14a^x'v  Sa^/. 

^?is.  25v'.x. 

-4ns.   4o^v  ac. 

^ns.  (a2_52)N/a. 

-4ns.  d(x  —  y)  \^xy, 

Ans.  Q>x(pxyy^. 

Ans.  2xy{^x'^zy\ 

Ans.  8m"(2a)^. 

Ans.  a^z(a  —  6)^. 


CASE  n. 

SOI,  To  reduce  a  rational  quantity  to  a  radical,  or  to 
introduce  a  coefficient  of  i  radical  under  the  radical  sisfn. 

1.  Reduce  bax^  to  the  form  of  the  cube  root. 

OPERATION. 
bax"  =  {pax?y  n=  V\2ba?x^ 
18 


Analysis.  "We  cuhc  each 
factor  of  the  given  quantity 
separately,  and  indicate  the 
cube  root  of  tbe  result 


J 


206  RADICAL  QUANTITIES. 

2.  Reduce  1c \' x  to  a  radical  without  a  coefficient. 

Analysis.     We  laise 
OPERATION.  the  coefficient,  2c,  tu  the 


(2c)^  =  (16c*)* 


fourth  power,  and  we 
have  (IGC*)*.  Multiply. 
ing  this  result  by  re* ,  we 


{\<6&y  X  x^  =  (IGc^o;)^ 

Hence         2c^ x  =   ^ V^d^x  have    V^lGc^x-.     Hence, 

the 

Rtile.  I.  To  reduce  a  rational  quantity  to  a  radical :  — 
Involve  it  to  the  same  power  as  the  requii^ed  index ^  and  write 
the  result  under  the  corresponding  radical  sign. 

II.  To  introduce  a  coefficient  of  a  radical  quantity  under 
the  radical: — Involve  it  to  the  same  power  as  the  radical, 
multiply  the  radical  by  the  result^  and  write  the  product 
under  the  radical  sign. 

EXAMPLES   FOR   PRACTICE. 

8.  Keduce  ax^z^  to  the  form  of  the  square  root. 

Ans.   -J  a?-xH^\  . 

4.  Keduce  9a^?/  to  the  form  of  the  cube  root. 

Ans.   VlWo>Yy  or  (729a^y)^. 

5.  Reduce  a  +  ex  to  the  form  of  the  fourth  root. 

Ans.   (a*  +  4a3cx  +  GaVx"  +  ^a&x""  +  c^x^Y . 

6.  Introduce  the  coefficient  oi  a^^/c  under  the  radical  sign. 

Ans.   "^  a^c. 

7.  Introduce  the  coefficient  of  3a >/2a*.;c  under  the  radical 
sign.  Ans.    ^  54a'a:. 

Reduce  the  following  quantities  to  equivalent  radicals  with* 
cut  coefficients  : 

8.  (2a  -  c)^^4.      Ans.   (32a»— 48a2c  +  24ac'— 4c«)K 

9.  4c' v^^  Ans.    ^T024ac^^ 
10.  ax^(a  +  bxyy,                           Ans.    ^  aV  +  a'bx'y. 


REDUCTION.  207 


11.  Cg6  +  x)yd'h-  —  'Zahx  +  x'. 


Ans.   s^^aW—  2d'h'x'  +  x^. 
12    (a^  —  ¥)s^.  Ans.  (a'  —  2a'h'  +  ah^y. 

CASE  m. 
20f5.  To  reduce  radicals  of  diflerent  degrees  to  a 
common  radical  index. 

1.  Reduce  a^  and  h^  to  a  common  radical  index. 

Analysis.   "We  have  seen  (201) 

OPERATION.  that  the  value  of  any  quantity  is 

i  3  not  changed  by  involving  it  to 

a    ■=  d  any  power  and  placing  the  result 

7)3  =  h^  under  the  corresponding  radical 

,  J      j^  ^  sign  or  index.     That  is,  the  nth 

a^  =  (a^)^,  b^  =  (&0^  root  of  the   nth   power   of   any 

Qx'  quantity  is   the    quantity  itself. 

I        g  i_        6  /  T-  Now,  as  the  index  of  a  in  the  given 

a^  =  V  a^,  6^  =  V  i''  example  is  I,  and  of  h  -J-,  we  may 

raise  the  two  quantities  to  any  powers  that  will  make  the  deno- 
minators of  their  indices  the  same.  This  we  do  by  reducing  the  in- 
dices -J  and  J  to  equivalent  indices  having  a  common  denominator, 
as  shown  in  the  operation.     Hence, 

KuLE.     I.  Reduce  the  indices  to  a  common  denominator. 

II.  Write  the  numerator  of  each  equivalent  index  as  an 
exponent  of  its  respective  quantity^  and  place  the  result  un^ 
der  the  common  radical  sign  or  index, 

EXAMPLES   FOR   PRACTICE. 

9         1  3 

2.  Reduce  a^,  d'\  and  cn  to  a  common  radical  index. 


OPERATION. 

1. 

h 

3  __ 

n  ~ 

4n  Sn 

18 
6n 

2 

a^ 

= 

An 

a6»i  = 

:  V^"  = 

=  (a^^L 

1 

dn 

= 

a67t  =rr 

VdF-  = 

=  {d'yn 

3 

r= 

18 

cen  = 

v;?  = 

=  Cc^«>l 

208  REDUCTION. 

8.  Keduce  m,  (an)^,  cxy'^^  and  5  to    a  common    radical 
index.  Ans,  (m«)^,  {ahi^^^^,  (c'xY)\  (125)* 

4.  Reduce  a*j?,  (any  and  v^cZ  to  a  common  radical  index. 

Ans,  'V  ^i^  V  7^,  'Vd\ 

ADDITION   OF   RADICALS. 
303.    1.  What  is  the  sum  of  3  s^ab  and  5  \^ab  ? 

OPERATION. 

o  ^~T  Analysis.     We  make   the   common  radical, 

y—  \/a6,  the  unit  of  addition ;  and  adding  the  co- 

efficients, we  obtain  8  \/a6,  the  required  sum. 


2.  What  is  the  sum  of   ^2bi)a'  and  >/lM? 

OPERATION.    Analysis.     We  reduce  the  radi- 

v^250a^=  5a  ^2a'^  ^^^^  *^   their  simplest    form,   and 

3 . — ; —                 3/- —  obtain  two  similar  radicals.     Add- 

vlba     =  -^      "<^  ing    their    coefficients,     we     have  . 

Sum,  7a  ^2^         7a  v'2^. 

From  these  examples  we  deduce  the  following 

Rule.     I.  Reduce  each  radical  to  its  simplest  form, 
II.  If  the  resulting  radicals  are  similar,  add  their  coeffi- 
dents,  and  to  the  sum  annex  the  common  radical ;  if  dis- 
similar, indicate  the  addition  by  the  plus  sign. 

EXAMPLES  FOR   PRACTICE. 

3.  Add  3  \/2>a^x  and  a  V'48a;  together      Ans.  7a  >/%x. 
4    Add  VSOm  and  Vl25m  together.        Ans.  9  v^5m. 

5.  Add  v/72,  n/I28,  and  ^\  Ans.  16v^2. 

6.  Add  xv/3^,  32/\/3a^,  and  2%/3^J^l 

Ans.  ^xy>/Za. 


7.  Find  the  sum  of  >/2a^xy  and  VWxy. 

Ans.  (a  4-  b)\^2xy. 


SUBTRACTION.  209 


8.  Find  the  sum  of  ^a'-x  —  d'y  and  \/'^d\x  —  y). 


An^.  Sav^x  —  y. 

9.  Find  the  sum  of  \/80^  and  v/245a«6^.  _ 

An^,  (4a6+7a^6^)v/5. 

10.  Find  the  sum  of  3V3a^^,  y/Vld^x,  and  v^36^a:. 

^ns.   (5a  4-  lS)V'6x, 

11.  Find  the  sum  of  >/2^2\/2^^,  and  v^2F. 

An$.   (a  4-  fe)'V2. 
12    Find  the  sum  of  V^^  and  l/Vlha\  _ 

Ans.  7Va\ 

13.  Find  the  sum  of  V270a'm  and  VlM0¥m7 

A71S,  (3a  +  66)Vl0^. 

14.  Find  the  sum  of  l^^%VSxY,  and  ^^ 

Ans.  (x+yy  ^ xy, 

15.  Find  the  sum  of  \ /  r-^  and  \ /  ~.  ^ris.   Va.   r—.. 

V    16  ^16 

16.  Find  the  sum  of  \/^  and  Vab'^ 

xins.  a\^b  +  b  Va. 

17.  Find  the  sum  of  \^a^i  and  V^. 

18.  Find  the  sum  of  2(Aa''by  and  (36tt^5)^". 

Ans.  10 ab-  '"^ 

19.  Find  the  sum  of  (aV  —  aVy)^  and  a(x*  —  o;^?/) ^     . -- 

^;is.  2ax(x  —  y)^, 

SUBTRACTION   OF  RADICALS.  / " 

S©4.  1.  From  \/98a  take  v/50a^ 

OPERATION.  Analysts.     Ileducin;^  the   radicals   to 

their  simplest  form,  we  obtain  the  two 

V  98a  =  7v2a  similar  radicals,  7  v^2^  and  5  V2S.  Making 

\/ ^Oa  =  5v^2a  the  radical  part  the  unit  of  subtraction, 

we  take  the  difference  of  the  coefficients, 

Difference,    2  V2a  and  obtain  2\/2a.    Hence  the  following 

18*  o 


210  RADICAL   QUANTITIES. 

Rule      I.  Beduce  each  radical  to  its  simplest  form. 

II.  If  the  resulting  radicals  are  similar,  subtract  the  coef- 
ficient of  the  subtrahend  from  the  coefficient  of  the  minuend^ 
and  to  the  remainder  annex  the  common  radical;  if  dissimi- 
lar ^  indicate  the  subtraction  by  the  minus  sign. 

EXAMPLES   rOR   PRACTICE. 

2.  From  3\/5a^c  take  aV^dc.  Ans.  2av^e. 

8.  From  V^162:c*2/ take  4\/8x^.  Ans.  x^s^^y. 

4.  Yrom  \^'A{)a%^ytfikQ'yba%y.  Ans.  a%^bhy. 

5.  From  3  v^l28a'6(?  take  4.as/T%^.  Ans.  4.a^2bc. 

6.  From  ^375?&  take  VUab^.     Ans.  {ba—2b)^^ab. 

7.  From  ^lQaW)^  take  2a(a56')\  Ans.  ^a\ab''Y. 
^-      8.  From  VZd'c  -\-  Qabc  +  Sb"c  take  V^126''c. 

Ans.  (a  —  6)v^8c-. 

9.  From  v^2a"^c^  take  ^2bc'^.      Ans,  ac^'Za  —  ^\^2b, 
\      10.  From  \/a^  —  a^  take  Vah'  —  b\  

-4ns.   {a — b)^a  —  h. 

11.  From  v/Ftake  v^f  Ans.   W'l, 

^  12.  From  6^82  take  Q'^^S.  Ajis.  lOv'i 

18.  From24/32^nake4v/^^      Ans.  ^a{^a'—^¥\ 

MULTIPLICATION  OF  RADICALS. 
CASE  L 

SO^.  To  multiply  radicals  of  the  same  degree. 

Since  the  root  of  a  quantity  composed  of  several  factors  is 
obtained  by  extracting  the  root  of  each  factor  separately 
{I§,'5),  we  have  (a6>-=  a^  _X  5=;_         ' 

Or,  by  radical  sign,   V ab  =  Va  x  \^b  ; 
Conversely,    Va  x    ^b    =  Vab. 
Hence,  if  we  consider  a  and  b  as  representing  any  two  quanti- 
ties, and  n  the  index  of  any  root,  we  have 

The  2yroduct  of  the  roots  of  any  two  quantities  is  equat,  to 
the  root  of  their  product. 


MU-LTIPLICATION.  211 

I.  Multiply  3a  v^i^  by  2  v/^. 

OPERATION.  ,  Analysis      Since  the  pro- 

duct   will    be    the   same,    m 

Ba^'^x  X   2%/^/  =  6a\^xy  whatever    order   the    factors 

are  taken,  we  inultiply  the 
coefficients  Za  and  2,  and  obtain  6a;  and  the  radical  parts  \/x  and 
v/'y,  and  obtain,  by  the  principle  enunciated  above,  y^xi/ ;  and  the 
entire  product  is   Q)a\/xy,     Hence  the  following 

Rule  .     I.  MultijAy  the  coefficients  together  for  the  coeffi' 
oient  of  the  product. 

II.  Multiply  the  quantities  in  the  radical  parts  together ^ 
and  place  the  product  under  the  common  radical  sign. 

III.  Reduce  the  entire  result  to  its  simpAest  form. 

EXAMPLES  FOR   PRACTICE. 

2.  Multiply  2a^^Sx  by  4v^  Ans.  8aV^3^ 

3.  Multiply  b^ac  by  ^^am.  Ans.  da^mc. 

4.  Multiply  3>/5^by  4n/20x.  A7is.  UOx^^'y^ 

5.  Multiply  2\/9x^  by  s/oxyz.  Ans.  6x\^yz. 

6.  Multiply  2 v^  14  by  3^4.  Ans.  12v'7". 

7.  Multiply  S>/ghj2y/S.  Ans.  18. 

8.  Multiply  3N/2by4v/8.  Ans.  48. 

9.  Multiply  V6  by  >/T507  Ans.  30. 
10.  Multiply  ^Ihjs/I.  Ans.  -}V3. 

+  11.  Multiply  a  +  ^bhj  '^b.  Ans.  a\^b  +  b. 

12.  Multiply  X  +  v^2/  ^7  ^  —  ^V-  -^^s*  ^^  —  V- 

13.  Multiply  v^m  +  v^n  bj  v^m —  \^n.    Ans.  m  —  n. 

14.  Multiply   v^a  +  >/c  by  \^a  +  v^c. 

4^  u4ns.  a  +  2v^ac  +  c 

15  Multiply  a(6/ by  c(c?)-.  Ans.  ac{bdy. 

16.  Multiply  2c(o!'bdf^  by  (3a&)^.  ^ns.  2ac(3ra)'l 

17.  Multiply  (x'^^/)^  ^7  (p^y^y*  ^^^-  ^V. 


212  RADICAL  QUANTITIES. 

CASE  n. 
306.  To  multiply  radicals  of  difierent  degrees. 

1.  What  is  the  product  of  a-  multiplied  by  6^  ? 

OPERATION.  Analysis.     Letting  P  represent  the  pro- 

1    \_  duct  of  the  given  quantities,  we  form  equa- 

Y  =  a  b^     (1)  |.j(3^  (;^)^     Squaring  both  members  we  have 

p2  _.  Q^^       /o)  (^)>  in  which  the  index  of  the  factor  a  is  1. 

pg 3T 2        3  Cubing  (2)  we  have  (3),  in  which  both  factora 

J  are  cleared  of  their  radical  indices  ;  and  ex- 

P  =  {a^b^y  (4)  tracting  the  sixth  root,  we  have  (4),  in  which 
the  product  is  under  a  common  index. 

BECOND  OPERATION.       ANALYSIS.     We  first  reduce  the  given  ra- 

1  3  dicals  to  equivalent  quantities  having  a  com- 

^"  ~  ^2  mon  radical  index,  by  (Case  III.,  Reduction), 

o  ^^  J  9<nd  then  multiplying  a^  by  6^  by  Case  I., 

o^h^  ==  (aWy  we  have  the  same  result  as  before.     Hence, 

IluLE.  I.  Eeduce  the  radical  parts  of  the  given  quanti^ 
ties  to  a  common  radical  index. 

II.  Multiply  the  rational  and  radical  parts  separately ^  as 
in  Case  L 

EXAMPLES  FOR  PRACTICE. 
1  1  5 

2.  Multiply  a~  by  a^.  Ans.  a®. 

3.  Multiply  6^  by  (150)i  Ans.  30. 
4..    4.  Give  the  product  of  v^-^,  multiplied  by  V|-  ?    Ans.  I  v^|T 

5.  Give  the  product  of  2^  multiplied  by  2*  ?     Ans.  v^l28 

6.  Kequired  the  product  of  {a  +  by  (a  +  ^)  • 

b 

Ans.  (a  +  hy, 
vY7.  Required  the  product  of  4>/a"x  c>bVd  -f  x, 

Ans.  12b'Va'\d  +  xf. 

8.  Multiply  Y  I  by  y  J  Ans.  S/^.. 


APPLICATIONS.  213 

DIVISION   OF  RADICALS.  '' ■ 

CASE  I. 

SO?.  To  divide  radicals  of  the  same  degree. 

Since  iLe  root  of  a  fraction,  or  of  the  quotient  of  one  quan- 
tity divided  by  another,  is  obtained  by  extracting  the  root  of 
e^ch  term  separately  (t §■«!),  we  have 


Kb)  =  1' 


6" 

Or  by  the  radical  sign,  \  /  -=  -— . 

Conversely,  y-^^  =  \J  -  ;  hence 

The  quotient  of  the  roots  of  two  quantities  is  equal  to  the 
root  of  their  quotient, 

1    Divide  6a^V^'?/ by  2a  >/;r. 

OPERATION.  Analysis.   Dividing  the  coefficients, 

we  have  3a  for  the  new  coefficient. 

(Sa^\^ xy  And,  by  the  principle  stated  above, 

\/xy  divided  by  '^^=\ /  " — =  v^2/ ' 
the  entire  quotient,  therefore,  is  oa\/ y.     Hence  the  following 

Rule.     I.  Divide  the-  coefficient  of  the  dividend  hy  the 

coefficient  of  the  divisor. 

TI.  Divide  the  quantity  in  the  radical  part  of  the  dividend 
by  the  quantity  in  the  radical  part  of  the  divisor ^  and  place 
the  quotient  under  the  common  radical  sign, 

III.  Prefix  the  former  quotient  to  the  latter j  and  reduce 
the  result  to  its  simplest  form. 


214  RADICAL  QUANTITIES. 

EXAMPLES   FOR  PRACTICE. 

2.  Divide  4\^a6c  by  \lV ac.  Ans.  2\/5. 

3.  Divide  ^\'Zba}'xy  by  v^5a^;r.  Ans,  [m^y. 

4.  Divide  2v/20iW  by  \/2m.  Ans.  2Qm>/m. 

5.  Divide  v/lGO  by  n/8.  .4?is.  2x^5". 

6.  Divide  v^M  by  \/6  u4ns.  3. 

7.  Divide  8 n/ 72  by  2 v/ 6.  Ans.  8n^3." 

8.  Divide  3\/l0  by  \/T5.  ^7?5.  \/6. 

9.  Divide  (a'b'cy  by  (a&)^.  Ans.  a(bcy . 

10.  Divide  12(xY)*  by  S(xy)K  Ans.  ixij(y")K 

11.  Divide  (a'b'cd')^  by  (a25cZ)K  u4ns.  d(b'c)K^ 

12.  Divide  ^^a"^  —  a*6  by  Va.  Ans.  aVl  —  a6.   '"t 

13.  Divide  ^a^  — &^by  ^oTT.  J^ns.   \^'a^'b 

1-1.  Divide  (a^  —  Jfy  by  (a  —  6)i  Ans.  (a  +  ^>)i 

15.  Divide  sy-l  by  \/^-  ^^s-  —  >/ab^.  t 

CASE   II. 

S©8.  To  divide  radicals  of  different  degrees. 
1.  What  is  the  quotient  of  (aby^  divided  by  a^  ? 

OPERATION  Analysis.    Letting  Q  represent  tho 

1  quotient,  we  have  equation  (i).     Jlais- 

/-x   (^'  bj  ^^^     ing  both  members  to  the  sixth  power, 

V  we   have  (2)  an   equation   without  a 

radical  index.     Dividing  a^b'^  by  €^ 

Q6  __  ^  ^  we  have  (3) ;  and  extracting  the  sixth 

<^'  root  and  reducing  the  index  of  b  to 

Q^^  z=z  b^  (3)     its  lowest  terms  we  have,  (4)  tho  ro- 

Q    =  &^  =  b^  (4)     ^^'^^^^  ^^^"^^- 


APPLICATIONS.  215 

SECOND  OPERATION.  Analysis.      We  first  re- 

X  3  ,  duce  the  dividend   and   di- 

a'^  =z  a    =  (a^^  visor  to  the  same  radical  in- 

(a^b^y'  -^  (a^y  =  b^  —  6*  dex,  by  (202),  and  then  di- 

viding as  in  Case  I,  vre  have 
tne  same  result  as  before.     Hence, 

KuLE.  I.  Reduce  the  radical  parts  of  the  dividend  and 
divisor  to  a  common  radical  index. 

II.  Divide  the  rational  and  radical  parts  separately,  as  i?% 
Case  L 

EXAMPLES  FOR   PRACTICE. 

2.  Divide  (ax-y  by  (xyy,  Ans.  y  — r  • 

_2_ 

3.  Divide  \^x  by  \^x,  Ans,  x'^^- 

4.  Divide  30  by  Vb.  Ans.  Wb. 

5.  Divide  lQx{a  +  cf  by  5(a  +  c)\     Ans.  2x(a  +  cy'\ 

6.  Divide  (p?  —  x"")  (m  +  2/)"  ^1  («  +  x)(m  +  yy, 

Ans.  (a  —  x)  (in  +  ^z)  "'"• 

7.  Divide  %/f  by  v/f.  Ans.  2^-^^^^. 

8.  Divide  -^-=.  by  —y=  .  Ans.  x-~.     - 

V  ax  ^x  '^d 

PRINCIPLES 

RELATING   TO   THE   APPLICATION   CP   INVOLUTION   AND 
EVOLUTION. 

^I^9,  The  application  of  Involution  to  the  solution  of 
radical  equations,  is  governed  by  the  principles  illustrated 
by  the  three  following  examples. 

].  Kaise  Va  to  the  second  power. 

Analysis.     Placino;  the  pro- 
OPERATION.  ^^^t   ^f  tj^g  .lg^^gj.g  ^^^^^   ^^^ 

>/a  X  v^a  =  v^o^  =  a  common  mdical  sign,  (205),  wo 

have  \/a^  —  a. 


216  RADICAL  QUANTITIES. 

2i  Raise  >^'a  +  6  to  the  second  power. 

OPERATION.  Analysis.   Since  the 

_  _  given  quantity  is  a  bi- 

(  \^a  +  by  =  a  +  2b  v^a  +  b^  nomial,  we  write  a,  the 

square  of  the  first  term ; 
26  %/«,  twice  the  product  of  the  two  terms ;  and  h'\  the  square  of  the 
second  term. 

3.  Kaise  \^a  +  "^b  to  the  second  power. 

OPERATION.  Analysis.   Since  the 

_  quantity  is  a  binomial, 

(  ^/a  +  \^by  =  a  +  2  \^ab  +  b        we  write  a,  the  square 

of  the  first  term ; 
2  \^ab,  twice  the  product  of  the  two  terms ;  and  b,  the  square  of  the 
second  term. 

These  three  examples  establish  the  following  principles  : 

I  If  a  radical  quantify  be  involved  to  a  power  cor- 
responding to  the  radical  index^  the  radical  sign  will  be 
removed.  (1.) 

II.  If  a  quantity  containing  both  radical  and  rational 
terms  be  raised  to  any  power,  the  radical  sign  will  not  be 
removed.  (2). 

III.  If  a  quantity  consisting  of  two  radical  terms  of  the 
second  degree  be  squared,  the  result  will  contain  but  a  single 
radical  term.  (3). 

SIO.  The  application  of  Evolution  to  the  solution  of  equa- 
tions above  the  first  degree,  is  governed  by  the  principles  illus- 
trated in  the  following  examples : 

1.  Extract  the  nth  root  of  a". 

OPERATION.  Analysis.    We  divide  the  exponent  of  the  power 

n/-^  =  a  ^^  ^^^®  given  quantity,  by  the  index  of  the  required 

root  (185),  and  obtain  1  for  the  exponent  of  the 
root,  which  is  omitted  in  the  written  result. 


SIMPLE  EQUATIONS.  217 

2.  Extract  the  nth.  root  of  a"  +  6. 

Analysis.  By  the  principles  of  involution,  every  power  of  a  mo- 
nomial consists  of  one  term  only ;  and  the  powers  of  a  binomial  con- 
sist of  at  least  three  terms.  And  since  a"  -f  6,  the  given  quantity, 
has  more  terms  than  any  power  of  a  monomial,  and  a  less  number 
of  terms  than  any  power  of  a  binomial,  it  cannot  be  a  perfect  power, 
and  we  therefore  indicate  its  root,  thus,  ^a"  +  b. 

3.  Extract  the  square  root  of  a^b^  +  2a^b'^  +  a'^b\ 

OPERATION.  Analysis.    "We  find  by 

\/^=  a'^b  (1)         ^^'^^^'  ^^^  ^^^  ^^^'  ^^^^  *^^ 

y— —  ol  the  given  terms  are  per- 

,  V  a'b^  =  ab  (2)         ^^^^  squares,  and  that  twice 

2x  a'b  Xab^  2a'b^  (3)         ^^e  product  of  their  square 

y/a'^b'^  +  2a^¥  +  a^b^  =  a^b  +  ab         roots  is  equal  to  the  other 

term  of  the  given  quan- 
tity (3).  This  answers  the  condition  of  a  binomial  square  (186),  and 
we  have  a^b  +  ab  for  the  required  root. 

The  principles  illustrated  by  these  three  examples  may  be 
stated  as  follows : 

I.  The  exponent  of  a  quantity  will  be  removed  by  extract* 
ing  the  root  whose  index  corresponds  to  tJie  exponent.  (1). 

II.  The  root  of  a  binomial  is  necessarily  a  surd^  and  a 
binomial  always  becomes  a  radical  by  evolution.  (2). 

III.  A  trinomial  is  a  perfect  square  when  two  of  its  terms 
are  perfect  squares  and  positive^  and  the  remaining  term  is 
twice  the  product  of  the  square  roots  of  the  others ,  and  either 
positive  or  negative.  (3). 

SIMPLE  EQUATIONS 
CONTAINING  RADICAL   QUANTITIES. 

211,     1.  Given  4  +  \/ic-— 3  =  7  to  find  the  va*..     ^  x. 

OPERATION.  Analysis.    We  first  tra<>spose  4 

to  the  second  member,  so  that  the 
radical  may  stand  alone,  and  ob- 
tain (2).      We    next  square  both 


4,+  s/x  —  S  =  7     (1) 


V' j;  —  3  =  8  (2) 

X  —  3  =  9  (3)  members  to  clear  the  equation  of 

a:  =  12  (4)  the  radical  sign  (209, 1),  and  ob- 

19  tain  (8).  Keducing^,  wehave»  =  12. 


218  RADICAL  QUANTITIES. 

2.  Given  s/ x  —  2  +  \/ x  +  6  =  4  to  find  x, 

OPERATION.  Analysis.     In  order  to 

x/ JZ2  +  n/^T6  =  4  (1)         ^^°'f  *•>*  involution  of  x 

/ y ^  '  to  the  second  power,  we 

Vx-^1^\  —  N/^_+_b  (2)  transpose  one  of  the  radi- 

ir~2=16— .8v/a;+6  +  iC  +  6     (3)         cals  to  the  second  mem- 

\/^-|.d==3  (4)         ber,  and  obtain   (2).     In- 

ic-f-6=:9  (5)         volving  both  members  to 

^_.3  /gx         the  second  power,  we  have 

(3),  an  equation  containing 
cnly  one  radical.  Transposing  and  reducing,  we  have  (4),  in  which 
the  radical  stands  alone.  Squaring  both  members,  we  obtain  (5),  and 
reducing,  cc  =  3. 

From  these  examples,  we  derive  the  following 

Rule.  I.  If  the  equation  have  hut  one  radical  term  con- 
taining the  unknown  quantity,  transpose  the  terms  so  as  to 
make  the  radical  stand  alone,  as  one  member  ;  then  clear  the 
equation  of  the  radical  sign  by  involution,  and  reduce  as 
usual. 

II.  If  there  be  two  or  more  radical  terms,  clear  the 
equation  of  radicals  by  successive  involutions,  adjusting  the 
terms  at  each  step  according  to  the  principles  enunciated  in 
(209). 

Note.  1.  Equations  which  appear  to  be  higher  than  the  first  degree, 
sometimes  become  simple  equations  by  reduction  of  terms. 

EXAlVrPLES   FOR  PRACTICE. 

3.  Given  \^x  +  5  =  9,  to  find  x,  Ans.  x  =  16. 

. 4 

4.  Given  v  ^  —  3  =  "y         -,  to  find  x.     Ans.  a;  =  7. 

▼  X     ■  o 

5.  Given  ^^4  +  (a?  —  2) ^  =  3,  to  find  x,  Ans.  x  ==  27. 

6.  Given  x  —  -^x^  +  6  —  —  2,  to  find  x.  Ans.  x==i. 

7.  Given  x  +  v'a?^  —  7  =  7,  to  find  x.  Ans.  a?  =  4. 
^  Given  ^^x  +  12  =  2  -f  y^x,  to  find  x.  Ans.  x  =  4. 


SIMPLE  EQUATIONS.  219 

9    Given  2  +  (Sxy  =  v/5^  +  4,  to  find  x, 

A71S.  X  =  12. 

10.  Given  a;  +  2  =  ^4  +  a7V^64  +  ^to  find  .r. 

Ans,  a;  =  6. 

11.  Given  a;  —  ^  v^a?  =  v^a;^  —  x^  to  find  a?. 

Ans,  a;  =  ^|. 

12.  Given  Va;  — 32  =  16  —  v^^,  to  find  a;. 

Note.     2.  For  brevity,  put  a  =  16,  and  restore  the  value  of  a  in  tha 
inal,  or  reduced  equation. 

Ans,  X  =  81. 


13.  Given  \/3  +  a;  =    y.,   ,     ,  to  find  x.     Ans,  a?  =  3 
Vd  4-  X 


_6_ 

14.  Given  Vx  —  16  =  8  —  Vx,  to  find  a;. 

Ans,  X  =  25, 

15.  Given  Va;  +  8a  =  2  -v^a,  to  find  x,       Ans,  x  =  a. 

. 2a  a 

16.  Given  ^cx  +  a  =  ~v^,  to  find  x,        Ans.  x  =  ~» 

Vza  c 

17.  Given  \^x  +  2a  =  Vl^+  >/x  —  2a,  to  find  x. 

Ans,    cc  =  — , 
A 

1  m« 

18.  Given     .-„    -    =     .-- — =■  —  a:,  to  find  x. 


Ans.  aj=-/m^  —  1. 

v^^+28      -^^+38 

19.  Given  -7= =  -p= ,  to  find  x, 

Va;+    4      va;+    6 

Note. — Place  y/x  =  i/j  then  find  y 

-4w«    aj  =  4. 

2a 


SO    Given  ^a;+>/ci  +  ic  = 


(a  +  x)i* 


220  QUADKATIC  EQUATIONS. 

SECTION  IV. 

QUADRATIC   EQUATIONS. 

^13.  A  ftuadratic  Equation  is  an  equation  of  the  second 
degree,  or  one  which  contains  the  second  power  of  the  un- 
known quantity ;  as  ic^  =  9,  or  x^  +  3x  =  a.  Quadratic  equa- 
tions are  divided  into  two  classes,  pure  and  affected. 

31 3*  A  Pure  Quadratic  Equation  is  one  which  contains 
the  second  power  only,  of  the  unknown  quantity ;  as,  x^  =  25, 
or  x^  -f-  8a6  =  2c. 

jloTE. — A  pure  equation,  in  general,  is  an  equation  whicli  contains  only 
one  power  of  the  unknown  quantity. 

914.  An  Affected  Quadratic  Equation  is  one  which  con- 
tains both  the  second  and  the  Jlrst  powers  of  the  unknown 
quantity ;  as,  x^  +  Sx  =  10. 

215.  The  Eoot  of  an  equation  is  such  a  value  as,  when 
substituted  for  the  unknown  quantity,  will  satisfy  the  equation. 

PURE   QUADRATICS. 

S16.  Since  a  pure  quadratic  equation  contains  only  the 
second  power  of  the  unknown  quantity,  the  unknown  terms 
may  always  be  united  into  one  by  miking  the  unknown  quan- 
tity the  unit  of  addition.     Hence, 

Uvery  pure  equation  of  the  second  degree  can  he  reduced 
to  the  form  of  ax^  =  b]  in  which  a  and  b  are  supposed  to 
represent  any  quantities  whatever.     Thus  the  equation 

3^2  —  21  =  7  —  x\ 

becomes,  by  transposing  and  uniting  terms, 

4x^  =  28. 


^    .^^U^^^4^^,+■  Z^    h   Ir,    uH^  LiAJ^ 
Z.  <^«iA^<^   ^7w^'-  +  2(3^  =  6«,'^+4i,    t?  Z-W^  J^  . 


^^1 


ir.  VuhJL.  Lu/  fjL.  0:t<AlctLt fvhiniU^  iLfiUvAM^ni.  tj^fJuu 


I 


PURE  QUADRATICS. 


16.  Given  ^ x  —  8  =  — ; -,  to  find  x. 


223 


>/x-\- 


An^.  X  =  zizl. 


17.  Given  x  >/a^  +  x'  =  a^  —  x^,  to  find  x.  _ 

Ans,  x=  dhaV^. 

18.  Given  ^x^  —  a^  =  a  v^m  —  1,  to  find  x.  _^ 

Arts,  ic  =  zb  a  Vw. 


PROBLEMS 
PRODUCING   PURE   QUADRATIC   EQUATIONS. 

^IT.  A  problem  may  often  furnish  either  a  pure  or  an 
affected  quadratic  equation,  according  to  the  notation  assumed. 

1.  Find  two  numbers  whose  difference  is  6,  and  whose  pro- 
duct is  40. 

SOLUTION. 


Let  re  —  3  =  the  less  number ; 
ic  +  3  =  the  greater. 


o:^ 

—  9 

= 

40 

a) 

x' 

= 

49 

(^) 

X 

= 

7 

(3) 

X   - 

—  3 

= 

4, 

the  less  number ; 

■x 

+  3 

= 

10, 

the  £?r( 

?ater. 

Analysis.  We  repre- 
sent the  less  number  by 
X  — '3,  and  the  greater 
by  a;  4-  3,  thus  making 
the  difference  of  the  num- 
bers 6,  according  to  the 
first  condition  of  the  prob- 
lem. Multiplying  x  —  3 
by  a;  +  3,  we  obtain  x^ 
—  9,  the  product  of  the 
two  numbers,  which  we 
put  equal  to  40,  according  to  the  second  condition  of  the  problem, 
and  we  obtain  (i),  a  pure  equation,  lleducing,  we  find  a;  =  7,  if  we 
Uijo  only  the  plus  sign.  Then  x  —  3  =  4,  the  less  ;  and  oj  +  3  =  10 
the  greater. 

NcTR. — If  we  take  a;  =  — 7,  in  the  aoove  problem,  we  shall  have  z  —  3 
~:  —  10,  the  less  number,  and  x  -\-  S  =  —  4,  the  greater.  This  result, 
considered  in  an  algebraic  sense,  satisfies  the  conditions  of  the  problem, 
for  —4  —  (—10)  =r  6,  the  dilference,  and  (—4)  x  (—10)  =  40,  the 
product.  In  general,  however,  such  a  value  will  be  taken  for  the  ua- 
kuowu  quantity,  as  will  satisfy  the  conditions  arithmetically. 


//i 


224  QUADRATIC  EQUATIONS. 

This  problem  will  give  rise  to  an  affected  quadratic,  if  we 
assume  the  following  notation : 

Let  X   =  less  number, 

00  +  Q    =  greater, 
Then  x''+  Qx  =  40, 

an  equation  containing  both  powers  of  x^  and  for  the  solution 
of  which  rules  will  be  given  hereafter.  In  the  problems  which 
follow,  an  affected  quadratic  may  be  avoided  by  proper  no- 
tation. 

2.  The  sum  of  two  numbers  is  6,  and  the  sum  of  their  cubes 
is  72  ;  what  are  the  numbers  ? 

SOLUTION. 

Let  2x    =  difference  ; 

3+07    =  greater  number ; 
8  —  X    ==  less. 

(3  +  xy  =  27  +  27x  +    dx"  +  ar^,  cube  of  greater; 
(3  —  xy  =  27  —  27a;  +    9x^  —  ar^  cube  of  less. 
^  I  +18a;2=72  a) 

X^==l  (2) 

X    =1  (3) 

S  +  X  =  4f  greater  number ; 
3  —  x  =  2,  less. 

Analysis.  We  let  2x  represent  the  difference  of  the  two  numbers, 
3  4-ic,  half  the  sum  plus  half  the  difference,  the  greater  (153);  and 
3  —  Xf  half  the  sum  minus  half  the  difference,  the  less;  and  as  the 
Bum  of  these  quantities  is  6,  this  notation  satisfies  the  first  condition 
of  the  problem.  Cubing  each,  and  adding  the  results,  we  obtain  for 
the  sum  of  the  cubes,  54  +  18x^  which  we  put  equal  to  72,  accord- 
ing to  the  second  condition  of  the  problem.  Reducing,  we  obtain 
3  -f  a;  =  4,  the  greater  number  ;  and  3  — x  =  2,  the  less. 

3.  A  and  B  distributed  1200  dollars  each  among  a  certain 
number  of  persons.  A  relieved  40  persons  more  than  B,  and 
B  gave  to  each  individual  5  dollars  more  than  A  ;  how  many 
were  relieved  bv  A  and  B  ? 


PURE  QUADRATICS.  £25 


SOLUTION. 

Let  ^  +  20  =  th^  number  relieved  by  A. ; 
X  —  20  =  the  number  relieved  by  B. 

rru  ^-^^      .      ^         1200 

^^^^  ^-+20+    ^  =  ^=20      ''' 

240  240 

DmdiDg(i)by5    ^^^20+    ^  =  ^^120     <*' 

Assume  a  =  20  and  h  =  240 

Eq.  (2)  becomes     — ; —  +    1  = (s) 

x-\-a  X  —  a 

Keducing  (3)  x^  =  a(26  +  a)  (4) 

Restoring  values  of  a  and  6,  a-}  =  10000        (5) 

By  evolution  x  =  100  (6) 

'  a;  +  20  =  120,  number  A  relieved  ; 
Hence,    ' 


^>   |^_20== 


80,       "       B 


4.  Divide  tlie  number  56  into  two  such  parts^  that  their 
product  ehall  be  640.  Ans.  40  and  16. 

5.  Find  a  number,  such  that  one  third  of  it  multiplied  by 
one  fourth  of  it,  shall  produce  108.  An^,  86. 

6.  What  number  is  that,  whose  square  plus  18  is  equal  to 
half  its  square  plus  30^  ?  Ans,  5. 

7.  What  two  numbers  are  those,  which  are  to  each  other  as 
5  to  6,  and  the  difference  of  whose  squares  is  44  ? 

Note.  —  Let  6x  =  the  greater, ^nd  hx  =  tlie  less. 

An%,  10  and  12. 

8.  What  two  numbers  are  those,  which  are  to  each  other  as 
3  to  4,  and  the  difference  of  whose  squares  is  28  ? 

AuB.  6  and  8. 

9.  What  two  numbers  are  those,  whose  product  is  144.  and 
the  quotient  of  the  greater  divided  by  the  less  is  16  ? 

An^,  48  and  8. 
P 


226  QUADRATIC  EQUATIONS. 

10.  The  leDgth  of  a  certain  lot  of  land  is  to  its  breadth 
as  9  to  5,  and  its  contents  are  405  square  feet.  Kequired  the 
length  and  breadth  in  feet.  Aiis.  27  and  15. 

11.  What  two  numbers  are  those  whose  diiTerence  is  to  the 
greater  as  2  to  9,  and  the  difference  of  whose  squares  is  128  ? 

Ans.  18  and  14. 

12.  Find  two  numbers  in  the  proportion  of  ^  to  §,  the  sum 

of  whose  squares  shall  be  225  ? 

Note.  —  Multiplying  the  fractions  J  and  |  by  6,  or  reducing  them  to 
a  common  denominator,  we  find  their  ratio  to  be  3  to  4. 

Ans.  9  and  12. 

13.  There  is  a  rectangular  field  whose  breadth  is  |  of  the 
length.  After  laying  out  ^  of  the  whole  ground  for  a  garden, 
it  was  found  that  there  were  left  625  square  rods  for  mowing. 
Required  the  length  and  breadth  of  the  field. 

Ans.  Length,  80  rods;  breadth,  25. 

14.  Two  men  talking  of  their  ages,  one  said  that  he  was 
94  years  old.  *^  Then,"  replied  the  younger,  ^'  the  sum  of  your 
age  and  mine,  multiplied  by  the  difference  between  our  ages, 
will  produce  8512."     What  was  the  age  of  the  younger  ? 

Ajis,  18  years. 

15.  A  fisherman  being  asked  how  many  fish  he  had  caught, 
replied,  "  If  you  add  11  to  the  square  of  the  number,  9  times 
the  square  root  of  the  sum,  diminished  by  4,  will  equal  50." 
How  many  had  he  caught  ?  Ans.  5. 

16.  A  merchant  gains  in  trade  a  sum,  to  which  320  dollars 
bears  the  same  proportion  as  five  times  the  sum  does  to  2500 
dollars  ;  what  is  the  sum  ?  Aiis.  $400. 

17:  What  number  is  that,  the  fourth  part  of  whose  square 
being  subtracted  from  8,  leaves  a  remainder  of  4  ? 

Ans.  4. 

18.  There  is  a  stack  of  hay,  whose  length,  breadth  and  height 
are  to  each  other  as  the  numbers  5,  4  and  3.     It  is  worth  as 


PURE   QUADRATICS.  227 

many  cents  per  cubic  foot  as  it  is  feet  in  breadth ;  and  the 
whole  worth,  at  that  rate,  is  192  times  as  many  cents  as  there 
are  square  feet  in  the  bottom  of  the  stack.  Required  the 
dimensions  of  the  stack.  / 

SOLUTION. 

Let  5x  =  length ; 
4.r  =  breadth ; 
8x  =  height ; 
6x  X  4a;  X  ox  =  cubic  feet  in  stack ; 

5x  X  4x  =  square  feet  in  bottom  ; 
5a:  X  4a;  X  ox  X  4x  =  cost ; 
192  X  5x  X  4a;  =  cost. 


ba;  X  4x  X  3a;  X  4x  =  192  x  5a;  x  4a;      a) 

3a;  X  4a;  =  192  (2) 

a;^=  16  (3) 

a;  =  4  (4) 

Ans.  Length,  20  feet ;  breadth,  16 ;  height,  12. 

Analysis.  According  to  the  given  proportions,  we  let  5.x%  4x,  and 
3a;  represent  the  three  dimensions.  Then  bx  X  4x  X  3a;,  their  indi- 
cated product,  will  be  the  solid  contents ;  and  5a;  X  4a;,  the  area. 
According  to  the  conditions  of  the  problem,  we  multiply  the  cubic 
contents  by  4x,  the  breadth,  and  the  square  contents  of  the  bottom 
by  192,  and  obtain  two  values  for  the  cost,  which  being  put  equal  to 
each  other,  give  (i).  canceling  the  factors  5a;  and  4a;  from  both 
members,  we  have  (2).  Again  canceling  3  X  4,  or  12,  from  both 
members,  we  have  (3),  which  gives  a;  =  4.  Hence,  5a;  =  20  feet,  4a? 
=  16,  and  3a;  =  12,  the  required  dimensions. 

Note. — The  advantage  of  keeping  the  factors  separate,  as  in  the  solu- 
tion just  given,  has  been  fully  illustrated  in  the  former  part  of  the  book- 
The  pupil  may  apply  the  same  method  to  some  of  the  examples  which 
follow. 

19.  A  man  purchased  a  field,  the  length  of  which  was  to  its 

breadth  as  8  to  5.     The  number  of  dollars  paid  per  acre  was 

equal  to  the  number  of  rods  in  the  length  of  the  field :  and 

the  number  of  dollars  given  for  the  whole,  was  equal  to  13 

19* 


228  QUADRATIC  EQUATIONS. 

times  the  number   of  rods   round  the  field.     Required  the 
length  and  breadth  of  the  field. 

Arts.  Length,  104  rods  ;  breadth,  65. 

20.  There  are  three  numbers  in  the  proportion  of  2,  3,  and 
6  ;  and  their  product  is  equal  to  108  times  their  sum.  Re- 
quired the  numbers.  Ans.  12,  18,  and  30. 

21.  It  is  required  to  divide  the  number  of  14  into  two  such 
parts,  that  the  quotient  of  the  greater  divided  by  the  less,  may 
be  to  the  quoticLt  of  the  less  divided  by  the  greater,  as  16  i  9. 

Ans.  The  parts  are  8  and  6. 

22.  What  two  numbers  are  those  whose  sum  is  12,  and 
whose  product  is  35  ?  A7is.  7  and  5 

Note.  —  For  notation,  see  2d  problem. 

23.  The  difference  of  two  numbers  is  6,  and  the  sum  of  their 
squares  is  50  ;  what  are  the  numbers  ?  Ans.  7  and  1. 

24.  The  difference  of  two  numbers  is  8,  and  their  product  is 
240  ;  what  are  the  numbers  ?  Ans.  12  and  20. 

AFFECTED  QUADRATICS. 

218.  Since  an  affected  quadratic  equation  contains  both 
the  first  and  second  powers  of  the  unknown  quantity,  the  equa- 
tion will  contain  two  unknown  terms,  and  only  two,  after  the 
coefficients  of  each  power  are  united. 
Thus  the  equation, 

3x2  —  12x  =  180  —  x'  +  4x, 
becomes,  by  transposition,  and  reduction  of  terms, 

4x2  — 16^  =  180, 
Or,  x^  —   4x  =  45. 

^nd  if  we  represent  — 4  by  2a,  and  45  by  b,  we  have 
x^  +  2ax  =  b    (A).     Hence, 
Every  affected  quadratic  equation  can  be  reduced  to  the 
form  of  x^  +  2ax  =  6,  in  which  2a  and  b  are  supposed  to 
represent  any  quantities  whatever,  positive  or  negative. 


AFFECTED  QUADRATICS.  229 

SIO.  Since  the  first  member  of  the  general  equation  (a),  is 
a  binomial,  its  root  is  a  surd  (!BiO,  II),  and  the  equation  in  that 
form  cannot  be  reduced  by  evolution.  We  observe,  however, 
that  x^,  the  first  term  of  (A),  is  a  perfect  square,  and  2ax,  the 
second  term,  contains  x,  the  root  of  this  square ;  and  it  only 
requires  that  another  square  be  added,  such  that  twice  the  pro- 
duct of  the  two  roots  shall  be  equal  to  2ax,  the  second  term, 
to  constitute  this  member  a  perfect  square  (^10,  III).  The 
square  to  be  added  will  evidently  be  a\  giving 


x''  +  lax  +  a^  =  6  -\- o}     (b), 


1   nAnflR-  '^ 


in  which  the  first  member  is  a  perfect  square. 

But  a,  whose  square,  (a^),  we  have  added,  is  half  the  coeffi- 
cient of  X  in  the  second  term.     Hence, 

If  the  square  of  half  the  coefficient  of  the  first  power  of 
the  unknown  quantity  he  added  to  both  members  of  a  quad- 
ratic equation  in  the  form  of  x"^  -\-  2ax  =  5,  the  first  member 
will  become  a  perfect  square. 

Note.  —  The  term,  a^,  is  added  to  the  first  member  to  complete  the 
square;  and  to  the  second  member  to  preserve  the  equality.     (Ax.  1.) 


EXAMPLES  FOR  PEACTICE. 

Complete  the  square  in  each  of  the  following  equations : 
Ans.  ir^  +  4x  +  4  =  96  +  4. 
Ans,  x^  —  4x  +  4  =as  49. 
Ans.  x""  —  Ix  +  4„9  ^  8_i, 
Ans.  sc"  +  2x  +  lr=.  16. 
Ans,  x^  +  12x  +  36  =  64. 
Ans,  cc*  +  6x  +  9  =  25. 
Ans.  x"  —  15a;  +  2f  5  =  f 
Ans.  a;«  — |ic  +  i  =  ^g^ 


1. 

ir»  4-  4x  =  96. 

2. 

x«  —  4x  =  45. 

8. 

x'  —  lx=^  8. 

\. 

x'  +  2x  =  15. 

5. 

ic*  +  12a;  =  28. 

6 

a;«  +  6x  ==  16. 

7. 

a;^.— 15x  =  — 54. 

8. 

20 

230  QUiBRATIC  EQUiTIONS. 

9.  x'—^x^  J.  Ans.  x'^  —  -^  +  M-  =  tA' 

a  c  ,  o        «  a^        c     ,     a^ 

SS^.  Reducing  the  equation, 

x''+2ax-i-a''=:b+a\ 
By  evolution,  we  have  x-j-a  =ziz\^b+a^. 


Transposing  a,  x  = — a+\/6+a^  1st  root. 

Or,  X  = — a — >/6+a^,2d  root 

Hence, 

Every  affected  quadratic  equation  has  two  roots,  unequal 
in  numerical  value. 


find  the 


-,      r..            ,        4  — 32:c         8(1  +  ^)     ,     Tp 
1.    Given  a;2 . =  _X___ — L  j^  16, 

to 

values  of  x. 

OPERATION. 

.'-1 

_  32^  __  8(1  +  0.) 
3         ~         3         "^ 

a) 

3x»—    4 

+  32x  =    8  +  8x  +  48 

(2) 

^x" 

+  24:^  =  60 

(3) 

x" 

+     8:^;  =  20 

(4) 

x^  +  8x 

+  16    =  36 

(6) 

X 

-1-    4    =  ±6 

(6) 

x^2 

(7) 

Or, 

:c  =  _  10 

(8) 

Analysis.  Clearing  of  fractions,  we  obtain  (2).  Transposinpi;  and 
uniting,  we  have  (3).  Dividing  by  3,  we  have  (4),  an  equation  in  the 
form  of  (A).  As  8  is  the  coefficient  of  x^  we  add  16,  the  square  of  one 
half  of  8,  to  both  members,  and  obtain  (5),  in  which  the  first  member 
is  a  perfect  square.  Extracting  the  square  root  of  both  members,  we 
have  (6).  Transposing  4,  and  uniting  —  4  with  -f  6,  the  plus  value 
of  the  root,  we  have  .r  =  2,  the  first  value  of  ar;  uniting  the  —  4  with 


AFFECTED   QUADRATICS.  231 

<—  6,  the  mimig  value  of  the  root,  we  have  x  =  — 10,  the  second 
value  of  X.  Hence,  for  the  solution  of  an  afiected  quadratic  equa- 
tion, we  have  the  following  general 

KuLE.  I.  Beduce  the  given  equation  to  the  form  of 
x^  +  2ax  =  h. 

11.  Complete  the  square  of  the  first  member^  by  adding  to 
Loth  members  of  the  equation  the  square  of  one  half  the  co- 
efficient of  the  second  term, 

III.  Extract  the  square  root  of  both  members,  and  reduce 
the  resulting  equation, 

EXAMPLES   FOR   PRACTICE. 

Reduce  the  following  equations  : 

2.  x'  +  4x  =  96.  Ans.  a;  =  8,  or  — 12. 

3.  x^  —  4:x  =  45.  A71S.  a;  =  9,  or  —  5, 

4.  x^  —  7x  =  8.  Ans.  x  =  8,  or  —  1, 

5.  x^  '\-  2x  =  15.  Ans.  a;  =  3,  or  —  5. 

6.  x'  +  12;r  =  28.  Ans.  a?  =  2,  or  — 14. 

7.  x"^  +  Qx  ==i  16.  Ans.  cc  =  2,  or  —  8. 

8.  x"^  —  15;r  =  —  5^.  Ans.  ^  ==  9,  or  6. 

9.  ^2 __  1^  ^  i| 3,  ^r^s.  a:  =  7,  or  —  \/. 
10.  x'^  —  |x  =  J.  Ans.  ^  =  1,  or  —  J. 

Notes.  1.  The  ten  preceding  examples  are  all  in  the  form  of 
8;2  -|-  2az  =  b,  and  require  no  application  of  the  first  step  of  the  rule. 

2.  If  in  the  preparation  of  the  equations  which  follow,  the  square  of 
the  unknown  quantity  appears  with  the  minus  sign^  make  it  positive  bj 
changing  ail  the  signs  of  the  equation. 

12.  3^7^  --  25^7  =  —  72  +  bx.  Ans.  a;  =  6,  or  4. 
13    2x''  +  100  =  32^  — 10.                 Ans.  x  =  11,  or  5. 

14.  %x  —  300  =  204  —  ^x\  Ans.  x  =  12,  or  — 14. 

15.  bx^  +  80  =  —  505  —  llO^r. 

Ans.  X  =  —  9,  or  —  13. 

16.  2x^  —  9j7  =  —  4.  Ans.  a?  =  4,  or  ^, 


232  QUADKATIC  EQUATIONS. 


17.  1  —  .r  =  5 ~rc.  Ans.  ic  r=  4,  or  — 14. 

y.  nr.  -X.  A 


2x 76 

18.  3  —  x^  =  — ^ .  Ans.  a?  =  6,  or  —  5f . 

19.  x'  +  |  =  ^-J  +  if.  Ans.  a)  =  l,or-2i 

20.  J— 30  +  a;  =  2x  — 22.  Ans.  a;  =  8,  or— 4. 
^  21.  |  — I  4-  7i  =  8J.  Jns.  x  =  1|,  or  —  g. 

Aj^  22.  J_15  =  ^  — 14|.  ^ns.  cc  =  3,  or— J. 

2^-;^8-£l^-  ^n..  =  12,or_2. 

24.  cc  —  1  +  ——:  ==  0.  Jns.  ic  =  3.  or  2 

X  —  4 

^_    22  —  X       15  —  a?  .                or^        io 

25.  ^^      = ^.  ^ns.  cc  =  36,  or  12. 

20  X  —  6 

2x  —  7        cc  +  1  .                .             , 

26.  rr-  =  - — V-o.  %  Ans.  aj  =  4,  or  —  1. 

X  —  1        ^x  +  o 

27.  x^  +  2ax  =  3al  ^ns.  a;  ==  a,  or  —  3a 

28.  a;2  —  4cx  =  4cc?—  2d!x  —  3c'  —  d\ 

Ans.  x=Zc  —  dy  or  c  —  d. 

29.  x^  —  2aa;  =  m*  —  o?.  Ans.  x=  azhm. 

30.  x^  —  2cx  =  4m  —  c^  u4ns.  x  =  c  it  2  >/m. 


SECOND  METHOD  OF  COMPLETING  THE  SQUARE. 

3^1.  It  frequently  happens  in  the  reduction  of  a  quadratic 
to  the  form  of  x^  +  2ax  =  6,  that  some  or  all  of  the  terms 
become  fractional,  and  render  the  solution  complex  and  diffi- 
cult. In  such  cases,  it  will  be  sufficient  to  reduce  the  three 
parts  of  the  equation  to  the  simplest  entire  quantities,  by 
uniting  terms,  and  dividing  the  equation  by  the  greatest  common 


AFFECTED  QUADRATICS.  233 

divisor  of  the  two  members.     The  equation  will  then  be  in  the 
form  of 

ax^  -\-hx  =  c    (A), 

in  which  a,  5,  and  c,  are  entire  quantities,  prime  to  each  other. 

To  render  the  first  member  of  equation  (a),  a  binomial 
square,  we  may  make  its  first  term  a  perfect  square,  by  multi- 
plying the  equation  by  a,  and  afterwards  complete  the  square 
by  the  ezlQ  already  given.  The  operation  will  appear  as 
follows : 

ax^  +  bx    =  c]  (1) 

Multiplying  (i)  by  a,  aV  +  bax  =  ac ;  (2> 

Putting  y  =  ax  ] 

And  (2)  becomes  y^  +  by  =  ca ;  (S) 

b^  b^ 

Completing  the  square,     ^'  +  ^2/  +  r      =  ^^  +  j-     W 

The  first  member  of  equation  (4)  is  a  binomial  square  ;  but 
one  of  the  terms  is  fractional,  a  condition  which  we  are  seek- 
ing to  avoid.  The  denominator  of  the  fraction  is  the  square 
number,  4  ;  and  if  the  equation  be  multiplied  by  4  to  clear  it 
of  fractions,  the  first  member  will  still  be  a  square,  because  it 
will  consist  of  square  factors.  Hence,  multiplying  the  equa- 
tion by  4,  we  obtain, 

4i/2  +  4:by    +b^  =  4:ca  +  b' ;  (S) 

Restoring  value  of  y^  4:a'^x^  +  4:abx  +  &^  =  4ca  +  b\    (6) 

Factoring  this  result,  and  comparing  it  with  the  primitive 
equation^  ^.^us, 

Primitive  equation,        ax^  +  bx  =       c ;  (i) 

Square  completed,    4a(a^^  +  bx)  +  6^  =  4a(c)  +  b^.    (6) 

we  perceive  that  equation  (6)  may  be  obtained  by  mu-Itiplying 
equation  (i)  by  4a,  and  adding  b^,  to  both  members  of  the  result 
Hence, 

20* 


234  QUADRATIC  EQUATIONS. 

If  a  quadratic  equation  in  the  form  of  ax^  +  6^  =  c,  he 
multiplied  by  4  times  the  coefficient  of  the  second  power  of 
the  unknown  quantity^  and  the  square  of  the  coefficient  of  the 
first  power  be  added  to  both  sides,  the  first  member  will  be- 
come a  perfect  square. 

1.  Given  5x^  +  4xc  =  204,  to  find  the  values  of  x. 

OPERATION. 

bx"  +  4x  =  204  (1) 

lOOa;^  +  80x  +  16  =  4096  (2) 

10x+    4  =  ±64  (3) 

lOx  =  60  or  —  68  (4) 

x=:   6  or  —  6|  (5) 

Analysis.  To  complete  the  square,  we  multiply  equation  (i)  by 
4  times  5,  and  add  the  square  of  4  to  both  members,  and  obtain  (2). 
Extracting  the  square  root  of  both  members,  we  have  (3).  Trans- 
posing 4,  we  have  (4);  and  dividing  by  10,  we  obtain  a;  =  6,  or  —  6|. 

2.  Given  x^  —  5x  =  —  6,  to  find  the  values  of  x. 

OPERATION.  Analysis.     In   this  example, 

4  times  the  coefficient  of  x^  is  4: 
a;2  —  5x  =  —  6        0)      ^Yq  therefore  multiply  by  4,  and 
4x'  — (    )4-25  =  l  (2)      add   the   square   of   5   to   both 

2x  —  5  =  d=  1        (3)       members,    and   obtain  (2).     Re- 
jc  =  3  or  2    (4)      ducing  by  evolution  and  trans- 
position, we   have  a:  =  3,  or  2. 
As  the  second  term  of  a  binomial  square  is  dropped  in  extracting 
the  square  root,  we  may  place  (    )  in  the  equation,  preceded  by  the 
proper  sign,  when  we  complete  the  square. 

From  these  examples  we  derive  the  following 

KuLE.     I.  Reduce  the  equation  to  the  form  of  ax^  +  bx 

«=  c,  in  which  the  three  terms  are  entire,  and  prime  to  each 

other. 

II.  Multiply  the  equation  by  4  times  the  coefficient  of  x\ 
ard  add  the  square  of  the  coefficient  of  x  to  both  members. 

III.  Extract  the  square  root  of  both  members,  and  y^educe 

the  resulting  equation. 

NcTE. — If  the  coefficient  of  a:^  is  1,  the  second  method  will  be  applied 
"with  advantage,  provided  the  coefficient  of  x  is  odd;  but  if  it  is  even,  the 
first  rule  is  preferable. 


AFFECTED  QUADRATICS.  235 

EXAMPLES   FOR  PRACTICE. 

3.  Gi   ix»  2x  —  5x  =  117,  to  find  the  yalnes  of  x. 

Ans.  cc  =  9,  or  —  C^. 

4.  Given  Bx^ —  5x  =  28,  to  find  the  values  of  x, 

A71S.  x  =  4:j  or  —  |. 

6.  Given  2x^  —  a;  =  70,  to  find  the  values  of  x, 

A71S,  a:  =  5,  or  —  y. 
t)    Given  bx^  +  4x  =  273,  to  find  the  values  of  x. 

Ans.  cc  =  7,  or  —  7|. 

7.  Given  2x*  +  3ic  =  65,  to  find  the  values  of  x. 

Alls,  a:  =  5,  or  —  6^. 

8.  Given  Sx^  •{•  bx  =  42,  to  find  the  values  of  x. 

Ans.  a:  =  3,  or  —  4-|. 

9.  Given  8x*  —  7a;  4-  16  =  181,  to  find  x. 

Ans.  cc  =  5,  or  —  4|-. 

10.  Given  lOx*  —  8x  +  8  =  320,  to  find  x. 

Ans.  a  =  6,  or  —  5  J. 

11.  Given  Sx^  +  2x  =  4,  to  find  x. 

Ans.  x  =  —  ^±:^VW. 

12.  Given  bx^  4-  7x  =  7,  to  find  x. 

Ans.  a7=~/^±^3^\/2i. 

13.  Given  ~  +  j%==  ^£^,  to  findx.  Ans.  x  =  ^^'^^^^^ 

14.  Given  12  +  ic  =      ^^       ,  to  find  a;. 

^yi.s.  cc  =  5,  or  — 12. 
Find  the  approximate  roots  of  the  following  equations : 

15.  a^^  — 5x  =  — 2. 

Ans.  X  =  4.5615  +,  or  .4384  +. 

16.  2a;^_Sx==12. 

Ans.  X  =  3.3117  +,  or  — 1.8117  -f . 

17.  %x^  —  x^\. 

Ans.  X  =  .7675  +,  or  —  .4342  +. 


236  QUADRATIC  EQUATIONS. 

Ans.  X  =  1.6180  +,  or  —  .6180  +. 

19.  ^x'  +  3x  =  5. 

Ans.  X  =  .8042  +,  or  — 1.5542  +. 

20.  ic«  — 7a;  =  — 11. 

Ans.  X  =  4.6180  +,  or  2.3820  + 

HIGHER  EQUATIONS  IN  THE  QUADRATIC  FORM. 

Q22.  Any  equation  is  in  the  quadratic  form,  when  it  eon 
tains  but  two  powers  of  the  unknown  quantity,  and  the  index 
of  the  higher  power  is  twice  the  index  of  the  lower.  Such 
equations  are  reducible  to  one  of  the  following  forms : 

a^"  +  ax**  =  6  ;  or, 
ax^"  +  6x"  ==  c ; 

and  may  therefore  be  reduced  by  one  of  the  rules  for  quadratics. 

] .  Given  x^  —  4x*  =  621,  to  find  the  values  of  x. 

OPERATION.  Analysis.  To  sam- 

plify  the  application 

Put  y  =a^  of  the  rule  for  quad- 

y^  =  oc^  ratios,  we  put  y=a^ 

y^  —  ^y  =  621  m     ^°^    2/'  =  ^.      The 

•^  _  (    )    +  4    =  625  (2>     Siven  equation  then 

^  9 Hh  25  (3)     ^^^^^^  ^^^'  ^  quad- 

^         ;    I  27      or -23       (4)     '-«c  in  the  general 
^  form.  Solving  m  the 

or,  ar*  =  27      or  — ^     (5)     ^.^^^^    manner,    we 

a?    =  3        or  >/ — 23    (e)     obtain  (i).      Restor- 
ing the  ra^ue  of  y, 
we  have  (5)  a  pure  equation ;  and  extracting  the  cube  root  of  both 
members  we  have  a;  =  3  or  \/ —  23. 

Note  1.  It  will  be  remembered  that  the  odd  roots  of  a  negat»»'^  quan- 
tity are  realy  while  the  eveJi  roots  are  imaginary  (184).  Hence,  by  ex- 
tracting the  cube  root  of  —  23.  and  prefixing  the  minus  sign,  we  Snd  the 
approximate  value  of  the  secona  root  m  the  example  above.  Thus, 
^ir23  =  2.84+. 


2. 

Given  x*  - 

^x- 

=  56, 

to  find 

X, 

OPERATION. 

Put 
V- 

y 

y' 

=  .? 
=  a:^ 

15 
or  — 

.7 

y'  —  y 
(  )  +  i 

22/  — 1 

=  56 
=  225 

=  zfc: 

=  8 

(1/ 

(2) 
(3) 
(4) 

or, 

x'' 

=  8 

or  — 

.7 

(5) 

1 

=  2 

or(- 

■7) 

ff  (6) 

X 

=  4 

orr- 

•7) 

^(7) 

HIGHER  EQUATIONS  IN  THE  QUADRATIC  FORM.    937 


Analysis.  If  wo 
represent  x-  by  y,  cc* 
will  be  2/^  and  the 
equation,  (l),  takes 
the  usual  quadratic 
form.  Reducing  by 
the  second  rule,  we 
have  (4).  Restoring 
the  value  of  ?/,  we 
obtain  (5).  Extract- 
ing the  cube  root  of 
both  members,we  ob- 
tain (6);  and  squar- 
ing both  members  we 

N.  ^       -  2 

have  a;=4  or  ( — 7)^. 

NoTB    2.    The  expression  ( — 7)^  signifies  the  cube  root  of  the  second 
power  of  —  7 ;  or,  \/49  =  3.65  +. 

EXAMPLES  EOR  PRACTICE. 

3.  Given  x*  +  2x^  =  24,  to  find  the  values  of  x. 

OPERATION. 

aj*  -f  2x'  =  24  a) 

Completing  square,  a;*  +  (  )  +  1  =  25  (2) 

Extracting  root,  cc^  -f  1  =  db  5  (3) 

Transposing,  cc'  =  —  1  zfc  5  (4) 

Uniting  terms,  x*  =  4,  or  —  6  .  (5) 

Extracting  root,  x  =  it  2,  or  db  >/ —  6  (6) 

4    Given  a*  —  Zx^  =  550,  to  find  the  values  of  x. 


Ans.  a;  =  db  5,  or  db  ^—  22.  ] 

5.  Given  3x  —  x"^  =  44,  to  find  the  values  of  x.  \ 

Arts,  X  =  16,  or  13|- 

6.  Given  x*  —  7x'  =  8,  to  find  the  values  of  x.  1 

Ans,  X  =  2,  or  —  1,  \ 

7.  Given  x*  —  6x'  =  567,  to  find  the  values  of  x.  \ 

AriB.  X  =  8,  or  —  2.758  -f  -  i 


238  QUADRATIC  EQUATIONS. 

POLYNOMIALS  UNDER  THE  QUADRATIC  FORM. 

^S3.  When  a  polynomial  appears  under  different  pow<»-re 
or  fractional  exponents,  one  exponent  being  twice  the  other, 
we  may  represent  the  quantity  by  a  single  letter,  and  apply 
one  of  the  rules  for  quadratics,  as  in  the  last  article. 

1.  Given  (x^  +  2x)2  +  2(x'  +  2x)  =  80,  to  find  the  values 
of  X, 

OPERATION. 

Assume  y^  =  (x^  ^  2xy ; 

And  2/  =  ^^  +  2a; ; 

Then,  2/' +22/ =  80; 

Completing  square,  2/^  +  (  )   +    1  =  81; 
By  evolution,  y  +    1  =  =h  9  ; 

Reducing,  y  =  S^  ov  —  10. 

Restoring  the  value  of  y,  we  have  two  equations  containing  x ; 

Thus,  x^  +  2a?  =  8,         OTx'  +  2x      =  —  10  ; 

Whence,  a;^  +  2^  +  1  =  9,  or  ^'-^  +  2^  +  1     =  —  9  ; 

And,  O!   +  1    =db3;or^  +  l  =±  3^^^^; 

Therefore,  a;  =  2  or  —  4  ;         or,  a?  =  3\/ — 1 — 1 ; 

or,  --  (1  +  3  v/I=T). 

EXAMPLES  FOR  PRACTICE. 

2.  Given  (x  +  3)  +  2(x  +  3)'^  =  35,  to  find  the  values 
of  X, 

OPERATION. 

(a^  +  3)  +  2(a;  +  3)*  =  35  d) 

Completing  square,  (^  +  3)  +  (    )  +  1  =  36  (2) 

Extracting  root,  v'aj  +  3  +  l=:it6  (3) 

Transposing  and  uniting  terms^,  ^  x  +  S=^d  or  —  7  (*) 
Squaring  both  members,  a;  +  3  =  25  or  49    (5) 

Transposing  and  uniting,  ar  =  22  or  46    (6) 


FORMATION  OF  QUADRATIC  EQUATIONS.       £39 

3.  Given  (\j''  +  lyy  +  Kv''  +  ^2/)  =  96,  to  find  one  value 
of  y,  Ans,  y  =  2. 

4.  Given  10  +  cc  —  (10  +  a:)^  =  12,  to  find  one  value 
of  X,  Ans.  x  =  Q 

5.  Given  (-  +  2/)'+  (-  +  2/)  =  30,  to  find  y, 

Ans.  2/  =  3  or  2,  or  —  3  ±  n/3. 

6.  Given  (x  +  12)*  +  (x  +  12)^=  6,  to  find  the  valuea 
of  X,  Ans.  x  =  4:j  or  69. 

7.  Given  2x'  +  3a;  +  9  —  5v^2x^  +  3x  +  9  =  6,  to  find 
the  real  values  of  x. 

Ans,  x=S  OT  —  4  J,  or  a?  =  —  |  =b  |-  ^ — 55. 

8.  Given  (x  +  of  +  2b(x  +  a)*  =  Bb\  to  find  the  valuea 
of  X.  Ans,  x  =  b^  —  a,  or  816*  —  a. 

FORMATION  OF  QUADRATIC  EQUATIONS. 

S34.  The  Absolute  Term  of  an  equation  is  the  term  or 
quantity  which  does  not  contain  the  unknown  quantity. 

^^5.  The  roots  of  quadratics  possess  certain  properties 
which  enable  us  to  reconstruct  the  equation  when  its  roots  are 
known. 

Let  us  resume  the  general  equation, 

x'^+2ax=h ;  (a) 

Completing  the  square,  x^+2ax+a^=a^'\-b ; 
By  evolution,  x+a=±^  a^+b  ; 

Hence,  x= — a+\^a^+by  1st  root; 

And,  ir=— a— Va'+6,  2d  root. 

Adding  these  two  roots,  we  have 

—  a+  ^oT+l) 

—  a—Va'-i-b 
Sum,      T—  2a 


240  QUADRATIC  EQUATIONS. 

Multiplying  them  together,  we  have 


.  a  +    \^a'  +  b 


—  a—    \^o?  +  h 


a^ — aVo?  +  h 
+  a  Va^  +  h  —  {o?  +  h) 
Product,  —  h. 

If  we  transpose  the  absolute  term  of  equation  (a)  the  equa- 
tion will  appear  as  follows  : 

x^  +  2ax  —  6  =  0        (B) 

Comparing  the  sum  and  product  now  obtained,  we  con- 
clude that  in  every  equation  in  the  form  of  x^  +  2ax  —  b  =  0> 

I.  The  sum  of  the  two  roots  is  equal  to  the  coefficient  of 
the  unknown  quantity  in  the  second  tervfiy  taken  with  the 
contrary  sign, 

II.  The  product  of  the  two  roots  is  equal  to  the  absolute, 
term  taken  with  its  proper  sign. 

1.  Form  the  equation  whose  roots  are  4  and  — 12. 

OPERATION. 

Algebraic  sum  of  roots,  4  — 12  =  —  8 

Product  of  roots,  4  x  (  — 12)  ==  —  48 

Hence,  a;^^  8a?  — 48  =  0 

Analysis.  The  algebraic  sum  of  the  roots  is  —  8 ;  and  their  pro- 
duct is  —  48.  Hence,  from  property  (I),  we  have  8  for  the  coeffi 
cient  of  the  first  power  of  the  unknown  quantity ;  and  from  (II), 
we  have  —  48  for  the  absolute  term  in  the  first  member;  hence  the 
equation  is  x^  -^  Sx  —  48  =  0. 

From  these  principles  and  illustrations  we  have  the  following 

Rule.  I.  Write  the  second  power  of  the  unknown  quan- 
tity for  the  first  term. 


FORMATION  OF  QUADRATIC  EQUATIONS.      241 

II.  Take  the  algebraic  sum  of  the  roots,  with  its  sign 
changed,  as  the  coefficient  of  the  unknown  quantity  in  the 
secand  term. 

III.  Write  the  product  of  the  roots  with  its  proper  sign  for 
the  third  term,  and  place  the  whole  result  equal  to  zero 

EXAMPLES  FOR  PRACTICE. 

2.  Form  the  equation  whose  roots  are  10  and  —  7. 

Ans,  x^  —  3a;  —  70  =  0. 

3.  Form  the  equation  whose  roots  are  12  and  —  5. 

Ans.  «2~7x  — 60  =  0. 

4.  Form  the  equation  whose  roots  are  6  and  — 15. 

Ans.  a;2  +  9x  — 90  =  0 

5.  Form  the  equation  whose  roots  are  1  and  —  2. 

Ans.  ^2  +  X  —  2  =  0. 

6.  Form  the  equation  whose  roots  are  4  and  13. 

Ans.  x"  —  11 X  +  62  =  0. 

7.  Form  the  equation  whose  roots  are  —  5  and  —  3. 

Ans.  x^  +  8a;  +  15  =  0. 

8.  Form  the  equation  whose  roots  are  4^  and  5^. 

Ans.  x^  —  lOx  +  24.*75  =  0, 

SECOND   METHOD. 

SS6.  Let  us  suppose  that  a  and  h  represent  any  quantities, 
and  find  the  product  of  the  binomials. aj  —  a  and  a:  — ^  6 ;  thus, 

X  —  a 

X  —  b 


Product;  a:^  —  (a  +  h)x  +  ah. 

Now  by  placing  this  product  equal  to  zero,  thus 

x^  —  (a  +  h)x  +  a6  =  0, 

we  form  the  equation  whose  roots  are  a  and  h ;  because  the 
coefficient  of  a;  —  (a  +  6),  taken  with  the  contrary  sign,  is 
21  Q 


242  QUADRATIC  EQUATIONS. 

the  sum  of  a  and  6,  (S3«5,  I) ;  and  the  absolute  term,  a6, 
is  the  product  of  a  and  h  (S^^,  II).     Hence 

Every  quadratic  equation  in  the  form  of  x^  +  2.ax  —  h  =  ^ 
is  composed  of  two  binomial  factors,  of  which  the  first  term 
in  each  is  the  ujiknown  quantity j  and  the  second  term,  the 
two  roots  with  their  signs  changed. 

To  illustrate  this  by  a  numerical  example,  take  the  following 

Ciination : 

x^  +  4a^  —  60  =  0  (1) 

Transposing,  x^  -{-  4:x  ==:  60         (2) 

Completing  squp.re,  x^  +  (  )  +  4  =  64         (3) 

By  evolution,  x  4- 2  =  zb  8      (4) 

Yv^hence,  x  =  6,        1st  root ; 

And,  X  =  —  10,  2d  root. 

Connecting  these  roots  with  x,  with  their  signs  changed,  and 
multiplying,  we  have 

X  —  6 

X  +  10  . 

x"^ —  6x 

lOx  —  60 


x^+  4x  —  60  =  0 
Thus  we  have  reconstructed  equation  (i).     Hence, 

iluLE.  I.  Connect  each  root,  with  its  contrary  sign,  to  an 
unknown  quantity. 

II.  Multiply  together  the  binomial  factors  thus  formed^ 
and  place  the  product  equal  to  zero, 

EXAMPLES   EOR   PRACTICE. 

1.  Find  the  equation  which  has  3  and  —  2  for  its  roots. 

Ans.  x^  —  X  —  6  =  0. 

2.  Find  the  equation  which  has  5  and  —  9  for  its  roots. 

Ans.  x^  +  4x  —  45  =  0. 

3.  Find  the  equation  which  has  7  and  —  7  for  ilts  roots. 

Ans.  cf'  —  49  as  0. 


FACTORING  TRINOMIALS.  243 

4.  Find  the  equation  which  has  8  and  — 12  for  its  roots. 

Ans.  x^  -{-  4:x  —  96  =  0. 

5.  Find  the  equation  which  has  —  5  and  —  7  for  its  roots. 

Ans.  x"  +  12x  +  35  =  0. 

6.  Find  the  equation  which  has  a  and  a  —  6  for  its  roots. 

Ans,  x^  —  (2a  —  h)x  -f  a?  —  a6  =  0.     - 

7.  Find  the  equation  which  has  h  and  —  a  for  its  roots. 

Ans,  x^  +  (a  —  l))x  —  ah  =■  0. 

FACTORING  TRINOMIALS. 

^fM7*  The  principle  established  in  the  last  article,  enailes 
US  to  resolve  any  trinomial,  in  the  form  of  x^  -{-  ax  +  b,  into 
two  binomial  factors,  either  exact  or  approximate. 

I.  Resolve  cc^  -f  5x  +  6  into  two  binomial  factors. 

OPERATION.  Analysis.    Since  the  given 

a;2  _|.  5x   -f     6  =  —  0  quantity  is  in  the  form  of  a 

x^  4-  dx  = 6  quadratic   equation    reduced 

^2i/^\i95_.        1  *^  ^^^  member,  we  place   it 

^^^        ^ p^  __  _,    -t  equal   to  zero,  and  solve  the 

^x  4-     D  —  d=  1  resulting    equation.       Then, 

•'^  =  —  o,  or        A  changing   the    signs    of    the 

(x  4-  3)  (x  +  2),  factors.  roots,    and   connecting  them 

Or,   2  X  3  =-6,  absolute  term,  ^'^^^   ^'    ^'®   ^^^^    (^'  +  ^') 

2  +  3  =  5,  coefficient  of  x,  (^  +  ^),  the  factors  that  com^ 

And(     4- 3Wx  +  '^Wactors  pose  the  tr,nomiaU226).  Or, 

Ana  1^^  ^  6)  (^x+  ^),  tactois.  ^^  ^^^  ^^^   ^^  inspection, 

two  factors  of  6,  the  absolute  term,  whose  sum  is  equal  to  5,  the  co- 
efficient of  X  in  the  middle  term,  and  thus  form  the  binomial  factors 
sought.     Hence,  the  following 

FtULE.  I.  Place  the  trinomial  equal  to  zero,  and  reduce  the 
resulting  equation. 

II.  Connect  each  root,  with  its  sign  changed,  to  the  lowest 
flower  of  the  literal  quantity,  and  the  result  will  he  the 
binomial  factors  required.     Or, 

Find  by  inspection  two  factors  of  the  absolute  term,  whose 
hum  is  equal  to  the  coefficient  of  the  middle  term;  and  con* 
nect  each  ivith  its  proper  sign,  to  the  literal  quantity 


244 


QUADRATIC  EQUATIONS. 


EXAMPLES   FOR   PRACTICE. 


Factor  the  following  trinomials. 

2.  x'^^x  —  20. 

3.  a'  —  7a  +  12. 

4.  i^  —  7a  —  S. 

-18. 


-X' 


Ans,  (x  —  5)  (x  4-  4) 
Arts,  (a  —  3)  (a  —  4) 
A71S.  (a  — 8)  (a  +  1) 
Ans,  (x  —  6)  (cc  +  5} 
Ans.  (x  +  9)  (x  —  2) 
Ans.  (x  +  14)(ic  — 3) 

8.  x^  +  2x  —  5.      Ans.  (x  —  1.449  +  )  (x  +  3.449  +  ) 

9.  x^  —  Sx  +  8.        Ans.  (a; —  6.828  +  )  (^  —  1.172  +) 


x:' 

x^  +  7x- 

X' 


THE   FOUR  FORMS. 

928,  In  the   general  equation,  x^  +  2ax  =  b,  2a,  as  we 

have  seen,  may  be  either  positive  or  negative ;  and  b  msiy  be 

either  positive  or  negative  ;  therefore,  for  a  representation  of 

every  variety  of  quadratic  equations,  we  have  the  four  general 

formi; 

x^  +  2ax  =  6  (1) 

cc*  —  2ax  =  b  (2) 

x^  —  2ax  =  —  b     (3) 

x"^  -f  2ax  =  —  6      (4) 

A  reduction  of  these  several  equations  gives  for  the  values 
of  X  as  follows: 


x=  +  a±  Va"  +  b 
cc  =  +  a  db  V^a'-  —  b 
x  =  —  a  db  v^a^  —  b 


a) 

(2) 
(3) 
(4) 


REAL  AND  IMAGINARY  ROOTS. 

229.  By  examining  the  roots  of  the  four  forms  given  in 
the  last  article,  we  find  that  in  the  first  and  second  forms, 
fl^  -f  6  ,  the  quantity  under  the  radical  is  positive  ;  its  root  can, 


PROBLEMS.  245 

therefore,  always  be  taken,  either  exactly  or  approximately. 
But  iu  the  third  and  fourth  forms,  a^  —  6,  the  quantity  under 
the  radical  will  be  negative  when  the  term  h  is  greater  nume- 
rically than  d^ ;  in  which  case,  the  root  cannot  be  extracted 
(184),  and  must  be  imaginary.     Hence, 

I.  In  each  of  the  first  and  second  forms,  both  roots  are 
always  real. 

II.  In  each  of  the  third  and  fourth  forms,  both  roots  are 
imaginary  when  the  absolute  term  is  numericalli/  greater 
than  the  square  of  half  the  coefficient  of  the  unknown  quan- 
tity in  the  second  term. 

Note. — Imaginary  roots  of  an  equation  furnished  by  a  problem,  indi- 
cate that  the  conditions  of  the  problem  are  impossible  or  absurd. 

PROBLEMS 
PRODUCING   QVADRATIC   EQUATIONS. 

S30.  1.  If  four  times  the  square  of  a  certain  number  be 
diminished  by  twice  the  number,  it  will  leave  a  remainder  of 
80  ;  what  is  the  number?  Ans.  3. 

Note. — The  number  3  is  the  only  number  that  Tvill  answer  the  required 
conditions  ;  the  algebraic  expression  — 4  will  also  answer  the  conditions ; 
but  the  expression  is  not  a  number  in  an  arithmetical  sense. 

2.  A  person  purchased  a  number  of  horses  for  240  dollars. 
If  he  had  obtained  3  more  for  the  same  money,  each  horse 
would  have  cost  him  4  dollars  less ;  required  the  number  of 
horses.  Ans.  12. 

3.  A  grazier  bought  as  many  sheep  as  cost  him  240  doL 
lars,  and  after  reserving  15  out  of  the  number,  he  sold  the  rs- 
mainder  for  216  dollars,  and  gained  40  cents  a  head  on  the 
number  sold ;  how  many  sheep  did  he  purchase  ?     Ans.  75. 

4.  A  company  dining  at  a  house  of  entertainment,  had  to 
pay  3  dollars  50  cents ;  but  before  the  bill  was  presented 
two  of  them  had  left,  in  consequence  of  which,  those  who 
remained,  had  to  pay  each  20  cents  more  than  if  all  had  beej 
present ;  how  many  persons  dined  ?  Ans.  7. 

21* 


246  QUADRATIC  EQUATIONS. 

5.  There  is  a  certain  number,  which  beini:^  subtracted  from 
22,  and  the  remainder  multiplied  by  the  number,  the  product 
will  be  117  ;  what  is  the  number  ?  Ans.  13  or  9. 

6.  In  a  certain  number  of  hours  a  man  traveled  36  miles; 
if  he  had  traveled  one  mile  more  per  hour,  it  would  have  taken 
him  3  hours  less  to  perform  his  journey  ;  how  many  miles  did 
he  travel  per  hour  ?  Ans.  3  miles. 

7.  A  man  being  asked  how  much  money  he  had  in  his 
purse,  answered,  that  the  square  root  of  the  number  taken 
from  half  the  number  would  give  a  remainder  of  180  dollars  ; 
how  much  money  had  he  ?  Ans.  $400. 

8.  If  a  certain  number  be  increased  by  3,  and  the  square 
root  of  the  sum  be  added  to  the  number,  the  sum  will  be  17  ; 
what  is  the  number  ?  Ans.  13. 

9.  The  square  of  a  certain  number  and  11  times  the  num- 
ber make  80  ;  what  is  the  number  ?  Ans.  5  or— 16. 

10.  A  poulterer  going  to  market  to  buy  turkeys,  met  with 
four  flocks.  In  the  second  flock,  were  6  more  than  3  times  the 
square  root  of  double  the  number  in  the  first ;  the  third 
contained  3  times  as  many  as  the  first  and  second  ;  the  fourth 
contained  6  more  than  the  square  of  one  third  the  number  in 
the  third  ;  and  the  whole  number  was  1938.  How  many  were 
in  each  flock  ?  Ans.  18,  24,  126,  1770. 

Note. — Let  2x'^  equal  the  number  in  the  first.     Also  see  (223). 

11.  The  plate  of  a  mirror  18  inches  by  12,  is  to  be  set  in  a 
frame  of  uniform  width,  and  the  area  of  the  frame  is  to  be  equal 
to  that  of  the  glass ;  required  the  width  of  the  frame. 

A71S.  3  inches. 

12.  A  square  courtyard  has  a  rectangular  gravel  walk  round 
it  The  side  of  the  court  is  two  yards  less  than  six  times  the 
width  of  the  gravel  walk,  and  the  number  of  square  yards  in 
the  walk  exceeds  the  number  of  yards  in  the  periphery  of  the 
court  by  164  ;  required  the  area  of  the  court,  exclusive  of  the 
walk.  Ans.  256  yards. 


TWO  UNKNOWN  QUANTITIES.  247 

13.  A  and  B  start  at  the  same  time  to  travel  150  miles ;  A 
travels  3  miles  an  hour  faster  than  B,  and  finishes  his  journey- 
s' hours  l)efore  him  ;  at  what  rate  per  hour  does  each  travel  ? 

A71S.  9  and  6  miles  per  hour. 

14.  A  company  at  a  tavern  had  1  dollar  75  cents  to  pay  ; 
but  before  the  bill  was  paid  two  of  them  left,  when  those  who 
remained  had  each  10  cents  more  to  pay  ;  how  many  "/^ere  m 
ib3  company  at  first?  Ans.  7. 

15.  A  set  out  from  C,  toward  D,  and  traveled  7  miles  a  day. 
After  he  had  gone  32  miles,  B  set  out  from  D  toward  C,  and 
went  every  day  y*^  of  the  whole  journey ;  and  after  he  had 
traveled  as  many  days  as  he  went  miles  in  a  day,  he  met  A ; 
required  the  distance  from  C  to  D. 

Ans.  76  or  152  miles ;  either  number  will  answer  the  con- 
ditions. 

QUADRATIC  EQUATIONS 
CONTAINING   TWO   UNKNOWN   QUANTITIES. 

^31.  In  general,  two  equations  essentially  quadratic,  In- 
volving  two  unknown  quantities,  depend  for  their  solution  on 
a  resulting  equation  of  the  fourth  degree.  A  solution  may  be 
effected,  however,  by  the  rule  for  quadratics,  if  the  equations 
come  under  one  of  the  three  following  cases : 

1st.  When  one  of  the  equations  is  simple,  and  the  other 
quadratic. 

2d.  When  the  equati-ons  are  similar  in  form,  or  the  unknown 
quantities  are  involved  and  combined  in  a  similar  manner. 

3d.  When  the  equations  are  homogeneous. 

We  give  illustrations  of  the  three  classes  in  succession. 

1st.  Simple  and  Quadratic. 
1 .  Given  1^,  ^  2/  ==  S3  ]*  '  ^  ^""'^  "^  ""^^  ^ 


248  QUADRATIC   EQUATIONS. 


OPERATION. 


X  +2y  =.9  (I) 

^  x"  +  2y^  =  SS  (2) 

From  (1),  cc  =  9  —  2y  :z^ 

Sqnaring  (3),  a;^  =  81  —  S6y  +  ^f    (4) 

From  (2),  ic2  ^  38  —  2y^  (6) 

From  (4)  and  (5),  81  —  S6y  +  iy^  =  33  —  2y^  (6) 

Reducing,  y^  —  6?/  =  —  8  (7) 

"Whence,  2/  =  4  or  2  (8) 

And  from  (i),  a;  =  1  or  5  (a) 


2d.  Similar  Equations. 
2.  Given    |         /  ^  9i  |  >  to  find  a;  and  y, 


operation. 

X  ^-    2/  ==  10 

0) 

xy  =24 

(2) 

Squaring  a), 

x" 

+  2^:2/  +  2/^  =  100 

(3) 

Multiplying  (2)  by  4, 

4:xy           =  96 

(4) 

Taking  (4j  from  (3), 

x" 

—  2a;2/  +  2/'  =  4 

(5) 

Extracting  square  root  of  (5), 

X—    y  =  ±2 

(0) 

But  in  (1), 

x+    2/  =  10 

(7) 

Adding  (6)  to  G), 

2a?  =  12  or  8 

(SI 

Taking  (6)  from  (7), 

2y  =  8  or  12 

(0) 

Whence 

a;  =  6  or  4 

(10) 

And 

y  =  4  or  G 

(11) 

8.  Given    1  ^   i.  /i  _  rq  | »  *o  ^^^  ^  ^^^  2/- 


TWO    UNKNOWN   QUANTITIES.  £49 
OPERATION. 

X  +  y  =  10  a) 

x"  -\-  y^  =  58  (2) 


SquariQg  a),  nc^  +  2^2/  +  2/'  =  l^^  (2> 

Taking  (2)  from  (S),  2^2/  =  42  (4) 

Taking  (4)  from  (2),  a;^  —  2xy  +  y^  =  IQ  (6) 

Extracting  square  root  of  (5),  x  —  y   =  zt:  4:  «) 

But  in  (1),  ^  +  y  =10  (7) 

Whence,  ar  =  7  or  3         (8) 

And  2/  =  3  or  7         (9) 

4.  Given   |  ^  ;J;  ^3^  05 1 ,  to  find  x  and  y, 

OPERATION. 

X  +  y  =  d  a) 

^3^2/^==  35  (2) 

Cubing  (1),     rr*  +  3:^^?/  +  ^^y'  +  y'=12^  (3) 

Taking  (2)  from  (3),  ^x^'y  +  3x2/^  —  90  (4) 

Factoring  (4),  Sxy(x  +  y)=90  (s) 

Dividing  (5)  by  (i),  Bxy  =  18  (6) 

Or,  xy  =  6  (V) 

Combining  (i)  and  (7),  |  ic  =  3  or  2  (8) 

as  in  2d  example,   J  2/  =  2  or  3  (o)       . 

5.  Given    {  J,  Z  ^  Z  728  }  ^  ^^  ^""^  ^  ^""^  2/- 

OPERATION. 

CC  —  2/  =  8  (1) 

a:8^y3=,r728  (2) 

Dividing  (2)  by  a),         x^  -{-  xy  -i-  y^  =  91  (S) 

Squaring  (i),  cc^  —  2xy  +  ^7"  =  64  (4) 

Taking  (4)  from  (S),  3:7-2/  =  27  (5) 

Or,  xy  =  d  (e) 

Adding  (6)  to  (3),  x^  -f  2xy  +  i/'  =  100  CO 

Extracting  square  root  of  (7),  x  +  y  =  ±:10  (8) 

But  in  (1),  x>^y  =  S  (9) 

Whence,  'x==  9  or  —  1  (lo) 

And  2/  —  1  or  —  9  (H) 


250  QUADRATIC  EQUATIONS. 

8d.    HOMOGENEOUS    EQUATIONS, 

OPERATION. 

2x^  —    xy  =  Q  (Ij 

2y'  +  Sxy  =  8  (2) 

Assume  x  =  vy  (3) 

Substituting  vy  in  a)        2uy  —    vy^  ==  Q  (4) 

And  in  (2)  2y''  +  Bvy'  =  8  (6) 


From  (4)  y'  =  ^— 


2'i;=^ 


(6) 


From  (5)  y'-2-+^  <^> 

-,        .  ,  8  6 


jCiquauiiJg  (.";  tiuu  v/ 

2 

+ 

Sv          2v"^  —  V 

(«) 

Reducing  (9) 

8i;^ 

— 

•  ISv  =6 

(9) 

Whence, 

V     =:2 

(10) 

Substituting  v  in  (7) 

2/^-1 

(11) 

Whence, 

2/=:tl 

(12) 

From  (3)  and  (i2) 

a;  =r±2 

(13 

Note. — For  simplicity,  only  one  value  of -y  was  taken  in  equation  (10), 

S33.  Referring  to  the  three  classes,  we  find  that, 

1st.  When  the  equations  are  simple  and  quadratic^  they 
may  he  solved  by  ordinary  elimination. 

2d.  When  the  equations  are  similar,  they  may  he  solved  by 
taking  advantage  of  multiple  forms,  and  of  the  relations  ex- 
isting between  the  sum,  difference,  and  iDvoduct  of  the  un- 
known quantities. 

^d.  When  the  equations  are  homogeneous,  they  may  he 
solved  by  the  use  of  an  auxiliary  quantity. 


TWO  UNKNOWN  QUANTITIES.  251 

EXAMPLES   FOR  PRACTICE. 

Find  the  values  of  x  and  y  in  the  following  equations. 

(2x^y^l2\  I  X  =  7  or  — 11, 

^-  4x^  +  2^  =  53r  ^''^'    l2/=.2or— 31. 

(    X  —^  =    1\  .         j'a:  =  4or— -5, 

2-    1   .'_3j'  =  13|-  ^«*-    iy  =  lor_2. 

5  or  7, 


(       £C2/  =  35  3  ( 2/  =  /  or  5. 

rx_2/==-l|  r:«  =  6or-7, 

5     jo:  +2/  =1125)  r.  =  568, 


2/ 
fa:  — v=      4)  .  fa:  =  5or  —  1, 

i   X  —  2/  =  3  -  (2/  =  2or  —  5. 


AuB. 


2  or  3, 

3  or  2. 


Note.— Put  —  =  P,  and  ~  =  0. 


9. 


1^+2/^  =  1^2)  ^^^    j:.  =  5or3, 

U  4-y  =8     3  •  •    (2/  = 

|.«  +  ,«      .      ,  =  78)         ^^.    1^  =  5 

f2x'  — 3xy=50) 
"•    \    v^  —  2f  =50  3  • 

■■{ 


2/  =  3  or  5. 

9  or  3, 
3  or  9. 


.  cc  =  10or  — 5v^~.2, 


252  QUADRATIC  EQUATIONS. 

PROBLEMS 

PRODUCING   QUADRATICS   WITH   TWO   UNKNOWN   QUANTITIES. 

Note. — In  several  of  the  examples  which  follow,  the  pupil  may  have 
the  choice  of  using  two  symbols,  or  one,  in  the  solution.  It  will  be  use- 
ful to  solve  by  both  methods. 

S3S.  1.  Divide  100  into  two  such  parts,  that  the  sum  of 
their  square  roots  may  be  14.  Ans.  64  and  36. 

2.  Divide  the  number  14  into  two  such  parts,  that  the  sum 
of  the  squares  of  those  parts  shall  be  100.      Ans.  8  and  6. 

3.  Divide  the  number  a  into  two  such  parts,  that  the  sum 
of  the  squares  of  those  parts  shall  be  b. 

Ans,  i(a  ±  V26^ir^^. 

4.  It  is  required  to  divide  the  number  24  into  two  such 
parts,  that  their  product  may  be  equal  to  35  times  their  dif- 
ference. Ans,  10  and  14. 

5.  The  sum  of  two  numbers  is  8,  and  the  sum  of  their 
cubes  152 ;  what  arc  the  numbers  ?  Ans.  3  and  6. 

6.  Find  two  numbers,  such  that  the  less  may  be  to  the 
greater  as  the  greater  is  to  12,  and  that  the  sum  of  their  squares 
may  be  45.  Ans.  3  and  6. 

7.  What  two  numbers  are  those,  ^"^hose  difference  is  3,  and 
the  difference  of  their  cubes  189  ?  Ans.  3  and  6. 

8.  What  two  numbers  are  those,  whose  sum  is  5,  and  the 
sum  of  their  cubes  35  ?  Ans.  2  and  3. 

9.  A  merchant  has  a  piece  of  broadcloth  and  a  piece  of  silk.s 
The  number  of  yards  in  both  is  110  ;  and  if  the  square  of  the 
number  of  yards  of  silk  be  subtracted  from  80  times  the  num- 
ber of  yards  of  broadcloth,  the  difference  will  be  400.  How 
many  yards  are  there  in  each  piece  ? 

Ans,  60  of  silk ;  50  of  broadcloth. 
10.  A  is  4  years  older  than  B  ;  and  the  sum  of  the  square? 
of  their  ages  is  976.     What  are  their  ages  ? 

Ans.  A's  age,  24  years ;  B^s,  20  years 


TWO   UNKNOWN'   QUANTITIES.  9/33 

11.  Divide  tlie  number  10  into  two  such  parts,  that  the 
square  of  4  times  the  less  part  may  be  112  more  than  the 
square  of  2  times  the  greater.  Ans.  4  and  6. 

12.  Find  two  numbers,  such  that  the  sum  of  their  squares 
may  be  89,  and  their  sum  multipUed  by  the  greater  may  pro- 
duce 104.  A 118.  5  and  8. 

13.  What  number  is  that  which,  being  divided  by  the  sum 
of  its  two  digits,  the  quotient  6| ;  but  when  9  is  sub- 
tracted from  it,  there  remains  a  number  having  the  same  digits 
inverted  ?  Ans.  82. 

14.  Divide  20  into  three  parts,  such  that  the  continued  pro- 
duct of  all  three  may  be  270,  and  that  the  difference  of  the  first 
and  second  may  be  2  less  than  the  difference  of  the  second  and 
third.  Ans.  5,  6,  and  9. 

15.  A  regiment  of  soldiers,  consisting  of  1066  men,  forms 
into  two  squares,  one  of  which  has  four  men  more  in  a  side 
than  the  other.  What  number  of  men  are  in  a  side  of  each 
of  the  squares  ?  Ans.  21  and  25. 

16.  A  farmer  received  24  dollars  for  a  certain  quantity  of 
wheat,  and  an  equal  sum  for  a  quantity  of  barley,  but  at  a 
price  25  cents  less  by  the  bushel.  The  quantity  of  barley  ex- 
ceeded the  wheat  by  16  bushels.  How  many  busliels  were 
there  of  each  ? 

Ans.  32  bushels  of  wheat,  and  48  of  barley. 

17.  A  kborer  dug  two  trenches,  one  of  which  was  6  yards 
longer  than  the  other,  for  17  pounds  16  shillings,  and  the  dig- 
ging of  each  trench  cost  as  many  shillings  per  yard  as  there 
were  yards  in  its  length.     What  was  the  length  of  each  ? 

Ans.   10  and  16  yard^i. 

13.  A  and  B  set  out  from  two  towns  distant  from  each 

other  247  miles,  and  traveled  the  direct  road  till  they  met.     A 

went  9  miles  a  day,  and  the  number  of  days  at  the  end  of 

which  they  met,  was  greater,  by  3,  than  the  number  of  miles 

which  B  went  in  a  day.     How  many  miles  did  each  travel  ? 

Ans.  A,  1 17  miles ;  and  B,  130. 
22 


254  QUADRATIC   EQUATIONS. 

19.  The  fore  wheels  of  a  carriage  make  6  revolutions  more 
than  the  hind  wheels,  in  going  120  yards ;  but  if  the  circum- 
ference of  each  wheel  be  increased  1  yard,  the  fore  wheels  will 
make  only  4  revolutions  more  than  the  hind  wheels,  in  the 
same  distance ;  required  the  circumference  of  each  wheel. 

Ans.  4  and  5  yards. 

20.  There  are  two  numbers  whose  product  is  120.  If  2  bo 
added  to  the  less,  and  3  subtracted  from  the  greater,  the  pro- 
duct of  the  sum  and  remainder  will  also  be  120.  What  are 
the  numbers  ?  Ans.  15  and  8. 

21.  There  are  two  numbers,  the  sum  of  whose  squares  ex- 
ceeds twice  their  product,  by  4,  and  the  difference  of  their 
squares  exceeds  half  their  product,  by  4 ;  required  the  num- 
bers. Ans.  6  and  8. 

22.  What  two  numbers  are  those,  which  being  both  multi- 
plied by  27,  the  first  product  is  a  square,  and  the  second  the 
root  of  that  square  ;  but  being  both  multiplied  by  3,  the  first 
product  is  a  cube,  and  the  second  the  .root  of  that  cube  ? 

A71S.  243  and  3. 

23.  A  man  bought  a  horse,  which  he  sold,  after  some  time, 
for  24  dollars.  At  this  sale  he  lost  as  much  per  cent,  upon 
the  price  of  his  purchase  as  the  horse  cost  him.  What  did  he 
pay  for  the  horse  ? 

Ans.  He  paid  $60  or  $40;  the  problem  does  not  decide 
which  sum. 

24.  What  two  numbers  are  those  whose  product  is  equal  to 
the  difference  of  their  squares ;  and  the  greater  number  is  to 
the  less  as  3  to  2  ?  Ans.  Iso  such  numbers  exist. 

25.  What  two  numbers  are  those,  the  double  of  whose  pro- 
duct is  less  than  the  sum  of  their  squares  by  9,  and  half  of  their 
product  is  less  than  the  difference  of  their  squares  by  9  ? 

Ans.  The  numbers  are  9  and  12. 


ARITHMETICAL   PROGRESSION.  255 

SECTION  V. 

ARITHMETICAL  PROGRESSION. 

934,  An  Aritlimetical  Progression  is  a  series  of  numbers 
or  quantities,  increasing  or  decreasing  by  the  same  difference, 
from  term  to  term.  Thus,  2,  4,  6,  8, 10, 12,  <fcc.,  is  an  increas- 
ing or  ascending  arithmetical  series,  having  a  common  difference 
of  2 ;  and  20,  17,  14,  11,  8,  &c.,  is  a  decreasing  seiie3,  "ivhose 
common  difference  is  3. 

SS5,  The  Extremes  are  the  first  and  last  terms  cf  the 
series. 

236,  The  Means  are  the  intermediate  terms. 

CASE   I. 

Sar.  To  find  the  last  term. 

To  investigate  the  properties  of  an  arithmetical  progression, 
let  a  represent  the  first  term  of  a  series,  and  d  the  common 
difference.     Then 

a,  (a  +  d),  (a  +  2d:),  (a  +  3cZ),  (a  +  4cZ),  &c., 
represents  an  ascending  series  ;  and 

a,  (a  —  d),  (a  —  2d),  (a  —  8d),  (a  —  4(f),  &c., 
represents  a  descending  series.     And  we  observe, 

1st.    The  first  term,  a,  is  taken  once  in  every  term, 

2d.  The  coefficient  of  d  in  any  term  is  one  less  than  the 
number  of  the  term  counted  from  the  left.  Therefore  the 
tenth  te:m  would  be  expressed  by 

a  +  9d', 

The  ITth  term  by  a  +  IQd-y 

The  53d  term  by  o  +  52c?; 

The  ??th    term  by  a  +  (n  —  l)d 


256  ARITHMETICAL  PROGRESSION. 

When  the  series  is  descending,  the  sign  to  the  term  contain- 
ing d  will  be  minus,  the  20th  term,  for  example,  would  be 

a  —  lM 

The  nth  term  a  —  (n  —  l)d 

If  we  suppose  the  series  to  terminate  at  the  nth  term,  and 
represent  this  last  term  by  L,  we  shall  have  the  general 
formula, 

L  =  a  do(7i  —  l)d         (A) 

in  which  the  plus  sign  answers  to  an  increasing,  and  the  minus 
sign  to  a  decreasing  series.  Hence,  to  find  the  last  term,  we 
have  the  following 

E.ULE.  I.  Multiply  the  common  difference  by  the  number 
of  terms,  less  one. 

II.  Add  the  product  to  the  first  term,  when  the  progression 
is  an  increasing  series,  and  subtract  it  from  the  first  term 
when  the  progression  is  a  decreasing  series. 

EXAMPLES   FOR   PRACTICE. 

If  the  series  be  increasing, 

1.  When  a  =  2  and  <i  =  3,  what  is  the  tenth  term  ? 

Ans.  29. 

2.  When  a  =  3  and  J  ==  2,  what  is  the  12th  term  ? 

Ans.  25. 

3.  When  a  =  7  and  d  =  10,  what  is  the  21st  term  ? 

Ans,  207. 

4.  When  a  =  1  and  d  =  ^,  what  is  the  100th  term  ? 

Ans.  50^. 

5.  When  a  =  3  and  cZ  =  f ,  what  is  the  100th  term  ? 

Ans.  36, 

6.  When  a  =  0  and  d  =  ^,  what  is  the  89th  term  ? 

A?is.  11. 
If  the  series  be  decreasing, 

7.  When  a  ==  56  and  cZ  =  3,  what  is  the  15th  term  ? 

Ans.  14, 


ARITHMETICAL  PROGRESSION.  257 

8.  When  a  =  60  and  d  =  7,  what  is  the  9th  term  ? 

Ans,  4. 

9.  When  a  =  325  and  d  =  16,  what  is  the  13th  term  ? 

Ans.  133. 

10.  When  a  =  6  and  d  =  ^,  what  is  the  20th  term  ? 

Ans.  — 3^. 

11.  When  a  =  30  and  (^  =  3,  what  is  the  31st  term  ? 

Ans.  — 60. 

CASE  n. 

S38.     To  find  the  sum  of  the  series. 

It  is  manifest,  that  the  sum  of  the  terms  will  be  the  same, 
in  whatever  order  the  terms  are  written. 

Take,  for  instance,  the  series       3,     5,     7,     9,  11. 
And  the  same  inverted,  11,     9,     7,     5,     3. 

The  sums  of  the  terms  will  be     14,  14,  14,  14,  14. 

Take         a  a  +    cZ,    a  +  2cZ,    a  +  3^,    a  +  4{?^      (i) 

Inverted,  a  +  4cZ,    a  +  Bd^    a  +  2d,    a  +    J,    a 

Sums,      2a  +  4cZ,  2a  +  4:d,  2a  +  4d,  2a  -f  4d,  2a  4-  4d      (2) 

This  resulting  series  (2)  is  uniform ;  and  we  observe  that 
each  term  is  formed  by  adding  either  the  extremes  of  the  given 
series  (i),  or  terms  equally  distant  from  the  extremes  ;  and  that 
the  sum  of  its  terms  must  be  twice  that  of  the  given  series. 
Hence,  in  an  arithmetical  progression, 

I.  The  sum  of  the  extremes  is  equal  to  the  sum  of  any 
ether  two  terms  equally  distant  from  the  extremes. 

II  Tivice  the  sum  of  the  series  is  equal  to  the  sum  of  the 
extremes  taken  as  many  times  as  there  are  terms. 

If,  therefore,  S  represent  the  sum  of  a  series,  n  the  niimbei 
of  terms,  a  the  first  term,  and  iv  the  last  term,  we  shall  have 
2S  =  n{a  +  L) 

71 

Or,  S  =  ^(a  +  L)         (B) 

22*  B 


258  ARITHMETICAL   PROGRESSION. 

Hence,  tc  find  tlie  sum  of  the  series,  we  have  the  following 

"Rule.  IluUiply  the  sum  of  the  extremes  by  half  the  iiurn- 
her  of  terms, 

EXAMPLES  FOR   PRACTICE. 

1.  The  first  term  of  an  arithmetical  series  is  5,  the  last  term 
92.  and  the  number  of  terms  30;  what  is  the  sum  of  tte 
terms  ?  Ans.  1455. 

2.  The  first  term  of  an  arithmetical  series  is  2,  the  number 
of  terms  10,  and  the  last  term  30 ;  what  is  the  sum  of  the 
terms?  A7is.  160. 

3.  The  first  term  of  an  arithmetical  series  is  5,  the  number 
of  terms  85,  and  the  last  term  107 ;  what  is  the  sum  of  the 
terms?  Ajis.  1960. 

4.  The  first  term  of  an  arithmetical  series  is  7,  the  last  term 
207,  and  the  number  of  terms  21 ;  what  is  the  sum  of  the 
terms?  Ans,  2247. 

5.  The  first  term  of  an  arithmetical  series  is  6,  the  last  term  ■ 
—  3*-,  and  the  number  of  terms  20 ;  what  is  the  sum  of  the 
terms  ?  Ans.  25. 

GENERAL  APPLICATIONS. 

S39.  In  an  arithmetical  progression  there  are  five  parts, 
as  follows : 


1st.  The  first  term ; 

Symbol  a. 

2d.    The  common  difi*erence ; 

"      d. 

Sd.    The  number  of  terms ; 

"      n. 

4th.  The  last  term ; 

''     L. 

5th.  The  sum  of  the  terms ; 

"      S. 

The  formulas. 

1j--=  a  ±(n- 

-iH 

(A) 

S  ==  1  (a  +  L), 

(B) 

contain  all  of  the  parts  enumerated  above  ;  and  since  these 
equations  are  independent  of  each  other,  they  are  sufficient  to 


GENERAL   APPLICATIONS.  O59 

determine  any  tivo  of  the  parts  which  are  unknown^  provided 
the  other  three  are  known. 

1.  The  snm  of  an  arithmetical  series  is  1455,  the  first  terra 

6,  and  the  number  of  terms  30  ;  what  is  the  common  difference  ? 

Analysis.  The  example  give? 
1455,  5  and  3,  for  the  values  of 
S,  a  and  n  respectively.  ^Ve 
substitute  the  given  values  o^  a 
and  n  in  formula  (A)  and  obtain 
(D;  likewise,  substituting  the 
values  of  S  and  n  in  (B)^  gives  (2) 
We  thus  obtain  tv/o  equations 
with  two  unknown  quantities, 
d  and  L.  Reducing  (2),  gives 
(3)  from  which  we  obtain  L=92. 
Equating  the  two  values  of  L 
in  (1)  and  (4)  we  have  (5),  which 
reduced,  gives  cZ  =  3. 

Note.  —  If  only  the  last  term  had  been  required,  formula  (b)  would 
have  been  sufficient. 

2.  The  sum  of  an   arithmetical  series  is  567,  the  first  term 

7,  and  the  common  difference  2  ;  what  is  the  number  of  terms  ? 

OPERATION. 

S  =  567 
a  =  7 

d  =  2  Analysts.     Substituting  the 

f w~r  o  o  /ix   values  of  a  and  d  in  formula  (A), 

gives  (1),  and  the  values  of  S  and 
567  ---.  ^  (7  -f-  L)      (2)  «  in  formula  (B),  gives  (2).     Re< 
^  ducing  (1)  we  obtain  (3).     Sub- 


s 

OPERATION. 

s 

^  1455 

a 

=  5 

n 

=r80 

L 

=  5  4-  29^Z 

(1) 

1455 

=  15(5  +  L) 

•2) 

15L 

=  1380 

(3) 

L 

=  92 

(4) 

5  +  29^7 

—  92 

(5) 

2M 

=  87 

(<5) 

d 

=  3 

(T) 

L  =  5  -j-  2/1  (3)  stituting  this  value  of  L  in  ('-!). 

fl  we  obtain  (4),  which  cleared  of 

567  =  ,y  (12  -f  2n)  (4)  fractions,  gives  (5).    Completing 

a    ,    n  f;r»T  /c>  the    square   and   reducing,  we 

n'  4-  6n  =  567  (S)  t  ^' 

„H()+    9  =  576  <7l"^^Iir^'''  '"'''''"'- 

n  +    3  =  24  (7) 

n  =  21  (8) 


260  ARITHMETICAL  PROGRESSION. 

Hence,  for  the  determination  of  any  required  parts  in  an 
arithmetical  progression,  we  have  the  following 

HuLE.  Substitute,  in  the  formulas  (a)  aiid  (b),  the  known 
quantities  given  in  the  example,  and  reduce  the  resulting 
equations. 

PROBLEMS. 
8.  Find  seven  arithmetical  means  between  1  and  49. 
Note. — If  there  are  7  means,  there  must  be  9  terms;  hence, 
n  =  9,  a  =  1,  and 
L  =  49. 
Ans,  7,  13,  19,  25,  31,  37,  43. 
4.  The  first  term  of  an  arithmetical  series  is  1,  the  sum  of 
the  terms  280,  and  the  number  of  terms  32  ;  what  is  the  com- 
mon diderence,  and  what  the  last  term  ? 

Ans.  d=  \,'h  =  16^. 
6.  Insert  three  arithmetical  means  between  ^  and  -J. 

Ans.  The  means  are  f,  Z^,  ^\< 

6.  Insert  five  arithmetical  means  between  5  and  15. 

Ans.  The  means  are  6f ,  8^,  10,  llf,  l^. 

7.  Suppose  100  balls  be  placed  in  a  straight  line,  at  the 
distance  of  a  yard  from  each  other ;  how  far  must  a  person 
travel  to  bring  them  one  by  one  to  a  box  placed  at  the  dis- 
tance of  a  yard  from  the  first  ball  ? 

Ans.  5  miles  1300  yards. 

8.  A  speculator  bought  47  building  lots  in  a  certain  village, 
giving  10  dollars  for  the  first,  30  dollars  for  the  second,  50 
dollars  for  the  third,  and  so  on ;  what  did  he  pay  for  the 
whole  47  ?  Ans.  $22090. 

9.  In  gathering  up  a  certain  number  of  balls,  placed  on  the 
ground  in  a  straight  line,  at  the  distance  of  2  yards  from  each 
other,  the  first  being  placed  2  yards  from  the  box  in  which 
they  were  deposited,  a  man,  starting  from  the  box,  traveled 
11  miles  and  840  yards  ;  how  many  balls  weie  there  ? 

'ins.  100. 


PROBLEMS.  261 

10.  How  many  strokes  do  the  clocks  of  Venice,  which  go 
on  to  24  o'clock,  strike  in  a  day  ?  Ans.  300. 

11.  In  a  descending  arithmetical  series,  the  first  term  is 
730,  the  common  difference  2,  and  the  last  term  2  ;  what  is 
the  number  of  terms  ?  Ans.  365. 

12  The  sum  of  the  terms  of  an  arithmetical  series  is  280, 
the  first  term  1,  and  the  number  of  terms  32 ;  what  is  the 
common  difference  ?  A7is.  ^. 

13.  The  sum  of  the  terms  of  an  arithmetical  series  is  950, 
the  common  difference  3,  and  the  number  of  terms  25  ;  what 
is  the  first  term  ?  Ans.  2. 

14.  What  is  the  sum  of  n  terms  of  the  series  1,  2,  3,  4,  5, 

Ans.  S  =  7^(1  4-  n), 

Li 

PROBLEMS   IN  ARITHMETICAL    PROGRESSION 

TO  WHICH   THE   PRECEDING   FORMULAS,   (A)   AND    (B),   DO   NOT 
IMMEDIATELY   APPLY. 

^4^.  Since  the  sum  of  the  extremes,  in  an  arithmetical 
series,  is  equal  to  the  sum  of  any  two  terms  equally  distant 
from  the  extremes  (SS8,  I),  we  have  the  following  special 
properties : 

I.  When  three  quantities  are  in  arithmetical  progression^ 
the  mean  is  equal  to  half  the  sum  of  the  extremes, 

II.  When  four  quantities  are  in  arithmetical  progression, 
the  sum  of  the  means  is  equal  to  the  sum  of  the  extremes. 

Take,  for  example,  any  three  consecutive  terms  of  a  series, 

fliS 

a  +  2(i,  a  +  3c?,  ci  +  4cf, 

and  we  perceive,  by  inspection,  that  the  sum  of  the  extremes 
is  double  the  mean,  or  the  mean  is  half  the  sum  of  the  extremes. 


262 


ARITHMETICAL  PROGRESSION. 


.  Take  four  consecutive  terms, 

and  Tve  have  a+ 2d,  a+ -od,  a  +  id,  a  +  5d, 

(a  +  2d)  +  (a  +  dd)  ==  2a  +  7d,  sum  of  the  extremes: 
(a  +  od)  +  (a  +  M)  =  2a  +  7d,  sum  of  the  means 
To  facilitate  the  solution  of  problems,  when  three  terms  are 

in  question,  let  them  be  represented  by  (x  —  ?/),  x^  (x  +  y),  y 

l^eing  the  common  diiference. 

When  four  numbers  are  in  question,  let  them  be  represented 

by   (x  — 3?/),    (x  —  y),    (x+y),    (x  +  ^y),    2y    being    the 

common  difference. 

Such  notation  will  often  secure  the  formation  of  similar 

equations;  and  in  applying  the  principles  enunciated  on  the 

preceding  page,  the  common  difference  ivill  disappear  by 

addition. 

1.  Three  numbers  are  in  arithmetical  progression  ;  the  pro- 
duct of  the  first  and  second  is  15,  and  of  the  first  and  third  is 
21 ;  what  are  the  numbers  ? 

SOLUTION 

Let  y  =  common  difference, 

X  —  y  =  first  term, 
X  =  second  term, 

X  +  y  =  third  term.  Analysis.      We  represent 

the  common  difference  by  y, 
and  the  first  term  by  x — y. 
Then  a;  must  be  the  second 
term,  andar-fy,  the  third  term. 
From  the  given  conditions  we 
have  (1)  and  (2).  Dividing  (2) 
by  (1)  we  obtain  (3),  which  re- 
duced to  find  the  value  of  x, 
gives  (4),  Substituting  this 
value  of  X  in  (2),  we  have  (5), 
and  reducing  this  equation,  we 
find  y  =  2,  Hence,  from  (4), 
we  get  :c  =  5,  &c. 


x{x — y)  =  15 

(1) 

x'  —  y'  =  2l 

(2) 

x+  y       7 
ic     ~  5 

(3) 

(i) 

Y-y-  =' 

(5) 

2ly'=  21  X  4 

(6) 

2/'=  4 

(7) 

y=  2 

(8) 

X  =  5,  2d  term, 

(9) 

X  —  y  =  S,  l&t  term, 

(10) 

X  +  y=  7,  3d  term, 

(11) 

PROBLEMS.  263 

2.  There  are  four  numbers,  m  aritlimetical  progression ;  the 
Rum  of  the  two  means  is  25,  and  the  second  number,  multiplied 
by  the  common  difference  is  50 ;  what  are  the  numbers  ? 

Ans.  5,  10,  15,  and  20. 

3.  There  are  four  numbers,  in  arithmetical  progression  ;  the 
product  of  the  first  and  third  is  5,  and  of  the  second  and 
fourth  is  21 ;  what  are  the  numbers  ? 

Ans,  1,  3,  5,  and  7. 

4.  There  are  five  numbers,  in  arithmetical  progression  ;  the 
sum  of  these  numbers  is  65,  and  the  sum  of  their  squares  1005  ; 
-what  are  the  numbers  ? 

Note. — Let  x  =  the  middle  term,  and  y  the  common  difference.  Theu 
z  —  2y,  a;  —  y,  x^  z -\-  y,  z  -\-  2y,  will  represent  the  numbers. 

Ans,  5,  9,  13,  17,  and  21. 

5.  The  sum  of  three  numbers  in  arithmetical  progression  is 
15,  and  their  continued  product  is  105 ;  w^hat  are  the  num- 
bers ?  Ans.  3,  5,  and  7. 

6.  There  are  three  numbers,  in  arithmetical  progression; 
their  sum  is  18,  and  the  sum  of  their  squares  158  ;  what  are 
the  numbers  ?  Ans.  1,  6,  and  11. 

7.  Find  three  numbers,  in  arithmetical  progression,  such  that 
the  sum  of  their  squares  shall  be  56,  and  the  sum  arising  from 
adding  together  once  the  first  and  twice  the  second,  and  thrice 
the  third,  shall  amount  to  28.  Ayis.  2,  4,  6. 

8.  Find  three  numbers,  such  that  their  sum  may  be  12,  and 
the  sum  of  their  fourth  powers  962 ;  and  the  numbers  have 
equal  difi'erences  in  order  from  the  least  to  the  greatest. 

Ans.  3,  4,  5. 

9.  Find  three  numbers  having  equal  differences,  and  such 
that  the  square  of  the  least  number  added  to  the  product  of 
the  two  greater,  may  make  28,  but  the  square  of  the  greatest 
number  added  to  the  product  of  the  two  less,  may  make  44. 

Ans,  2,  4,  6. 


264  ARITHMETICAL    PROGRESSION. 

10.  Find  three  numbers,  in  arithmetical  progression,  such 
that  their  sum  shall  be  15,  and  the  sum  of  their  squares  93. 

Ans,  2,  5,  and  8. 

11.  Find  three  numbers,  in  arithmetical  progression,  such 
that  the  sum  of  the  first  and  third  shall  be  8,  and  the  sum  of 
the  squares  of  the  second  and  third  shall  be  52. 

Ans.  2,  4,  and  6. 

12.  Find  four  numbers,  in  arithmetical  progression,  such 
that  the  sum  of  the  first  and  fourth  shall  be  13,  and  the  dif- 
ference of  the  squares  of  the  two  means  shall  be  39. 

Ans.  2,  5,  8,  and  11. 

13.  Find  seven  numbers,  in  arithmetical  progression,  such 
that  the  sum  of  the  first  and  sixth  shall  be  14,  and  the  pro- 
duct of  the  third  and  fifth  shall  be  60. 

Ans.  2,  4,  6,  8,  10,  12,  and  14. 

14.  Find  five  numbers,  in  arithmetical  progression,  such 
that  their  sum  shall  be  25,  and  their  continued  product  945. 

Ans.  1,  3,  5,  7,  and  9. 

15.  Find  four  numbers,  in  arithmetical  progression,  such 
that  the  diflference  of  the  squares  of  the  first  and  second  shall 
be  12,  and  the  difference  of  the  squares  of  the  third  and  fourth 
shall  be  28.  Aiia.  2,  4,  6,  and  8. 


GEOMETRICAL  PROGRESSION.  265 


GEOMETRICAL   PROGRESSION. 

S4I*  A  Geometrical  Progression  is  a  series  of  numbers 
increasing  or  decreasing  by  a  constant  multiplier.  Thus  2, 
6.  18,  51,  162,  &c.,  is  a  geometrical  series,  in  which  the  first 
feerm  is  2,  and  the  multiplier  is  3. 

The  series  9,  3,  1,  |,  J,  ^V?  ^^-y  ^^  ^^so  a  geometrical  series, 
in  which  the  first  term  is  9,  and  the  multiplier  is  \. 

When  the  multiplier  is  greater  than  1,  the  series  is  ascend- 
ing,  and  when  the  multiplier  is  less  than  1,  the  series  is 
descending. 

^4^.  The  Ratio  is  the  constant  multiplier. 

Note.  —  The  words,  means  and  extremes,  as  defined  in  Aritlimetical 
Progression,  have  the  same  application  in  Geometrical  progression,  or  in 
any  series. 

CASE  I. 

S43.  To  find  the  last  term. 

If  we  represent  the  first  term  of  the  series  by  a,  and  the 
ratio  by  r,  then 

a,  ar,  ar^,  ar^,  ar*j  &c.,  \^ 

will  represent  the  series  ;  and  we  observe, 

1st.    The  first  term  is  a  factor  in  every  term. 

2d.  The  exponent  of  r  in  any  term  is  one  less  than  the 
number  of  the  term  ;  thus,  in  the  second  term  it  is  1 ;  in  the 
third  term,  2 ;  in  the  fourth  term,  3 ;  and  in  the  nth  term  it 
must  he  n  —  1. 

Therefore,  if  n  represent  the  number  of  terms  in  any  series, 
and  L  the  last  term,  then 

L  =  ar""-^        (A). 
23 


266                           GEOMETKICAL   PROGRESSION.  1 

Hence,  the  following  I 

Rule.     I.  Raise  the  ratio  to  a  power  whose  index  is  one  j 

less  than  the  number  of  terms.  \ 

II.  Multiply  the  result  by  the  first  term.                         '  | 

Note. — "^Ylien  any  two  successive  terms  in  the  series  are  given,  it  is  j 

evident  that  the  ratio  may  be  found  by  dividing  any  term  by  the  pre-  * 

ceding  term.  | 

EXAMPLES   FOR   PRACTICE.  3 

1.  The  first  term  of  a  geometrical  series  is  3,  and  the  ratio  \ 
is  2 ;  what  is  the  6th  term  of  the  series  ?  \ 

Ans.  L  ==  (2)s  X  3  =  9G.  \ 

2.  The  first  term  of  a  series  is  5,  and  the  ratio  4;  what  is  "\ 
the  9th  term  ?  | 

Ans.  327680.  j 

8.  The  first  term  of  a  series  is  2,  and  the  ratio  8 ;  what  is  \ 

the  8th  term  ?  i 

Ans.  4374.    •  j 

4.  The  first   term  of  a  series  is  1,  and  tlie  ratio  | ;  what  is  ] 
the  5th  term? 

Ans.     -ij\.  \ 

5.  The  first  term  of  a  series  is  2,  and  the  ratio  3 ;  what  is  \ 

the  7th  term  ?  \ 

Ans.  1458.  J 

6.  The  first  term  of  a  geometrical  series  is  5,  and  the  ratiG  j 

4 ;  what  is  the  6th  term  ?  I 

Ans.  5120.  1 
I 

7.  The  first  term  of  a  series  is  1  and  the  ratio  2 ;  what  is  \ 

the  10th  term  ?  \ 

Ans.  1024.  ] 

8.  The  first  term  of  a  series  is  1,  and  the  ratio  |  ;  what  is  j 

the  8th  term  ?  | 

Ans.  ri\%%.  I 


^  GBOMETRICAL   PROGRESSION.  267 

CASE   IL 

844.  To  find  the  sum  of  the  series. 

Let  S  represent  tlie  sum  of  any  geometrical  series  chen 
we  have 

S  =  a  +  r/r  4-  ar^  +  ar^^  &c.,  to  ar^^,      (l) 
Multiply  this  equation  by  r,  and  we  have 

rS  =  or  4-  or^  +  «^,  &c.,  to  ar'*"*  -f  ar^.      (£) 
Subtracting  (i)  from  (2),  we  have 

(r  —  1)S  =  ar'  —  a,      (») 
But  from  Case  I.  we  have 

L  =  ar^-\ 
And  multiplying  by  r, 

rL  =  ar". 
Therefore,  by  substituting  the  values  of  ai^  in  (s), 

(r  — l)S  =  rL  — a. 
Or.  S  =  -^^-y.     (B) 

Hence,  the  following 

Rule.  Multiply  the  last  term  by  the  ratio,  and  from  the 
product  subtract  the  first  term,  and  divide  the  remainder  by 
the  ratio  less  one. 

EXAMPLES   FOR   PRACTICE. 

1.  The  first  term  is  5,  the  last  term  1280,  and  the  ratio  4  5 
what  is  the  sum  of  the  series  ? 

Ans.  S  =  mj^A^'  =  1705. 

2.  The  first  term  is  2,  the  last  term  486,  and  the  ratio  8 ; 
what  is  the  sum  of  the  series  ?  Ans,  728. 


268                           GEOMETRICAL    PROGRESSION.  ^ 

3.  The  first  term  is  12,  the  last  term  7500,  and  the  ratio  5  ;  \ 

what  is  the  sum  of  the  series  ?                               Ans.  9372.  \ 

\ 

Note. — If  the  last  term  is  not  given,  it  may  first  be  found  by  Case  I.  \ 

\ 

4.  What  is  the  sum  of  8  terms  of  the  series,  2,  6,  18,  &c.  \ 

Ans,  05()0.  -l 

5.  What  is  the  sum  of  10  terms  of  the  series,  4, 12,  36,  &c.  ?  I 

Ans.  118096.  ] 

6.  What  is  the  sum  of  9  terms  of  the  series,  5,  20,  80,  (fee.  ?  ] 

Ans,  436905.  1 

7.  What  is  the  sum  of  5  terms  of  the  series,  3,  4i,  Q^,  &c.  ? 

Ans,  39 /g.  \ 

8.  What  is  the  sum  of  10  terms  of  the  series,  1,  f,  |,  <fec.  ?  1 

An(t       58025  \ 

^'^^'     1^68  3-  1 

9.  A  man  purchased  a  house,  giving  1  dollar  for  the  first  | 
door,  2  dollars  for  the  second,  4  dollars  for  the  third,  and  so  j 
on ;  what  did  the  house  cost  him,  there  being  10  doors  ?        •  \ 

Ans,  $1023.  j 

10.  A  farmer  planted  1  bushel  of  corn,  and  it  produced  20  \ 

bushels.     The  next  year  he  planted  the  20  bushels,  and  they 

produced  at  the  same  rate  as  the  first  bushel.     If  he  planted 

each  year's  crop  successively  for  5  years,  and  it  produced  at 

the  same  rate,  what  would  be   the   amount  of  the   5  years'  j 

harvest  ?  i 

j 

Note. — As  the  1  bushel  was  planted  and  the  20  bushels  were  harvested  \ 

the  first  year,  the  number  of  terms  is  6.  i 

Ans.  3368420  bushels.  \ 


INFINITE   SERIES. 


245»  By  formula  (B),  and  the  rule  subsequently  given,  we 
perceive  that  the  sum  of  the  series  depends  on  the  first  aiid  last 
terms  and  the  ratio.     Suppose,  now,  that  we  are  required  to  find 


INFINITE    SERIES.  269 

tlie  sum  of  a  descending  series,  as  1,  |,  |,  i,  &c.  We  perceive 
that  the  terms  decrease  in  value  as  the  series  advances  ;  the 
hundreth  term  would  be  extremely  small,  the  thousandth 
term  would  be  very  much  less,  and  the  infinite  term  noth- 
ing. 

Therefore,  as  L  becomes  0,  it  follows  that  in  any  decreasing 
series,  when  the  number  of  terras  is  conceived  to  be  infinite,  the 
sum  of  the  series  depends  wholly  on  the  first  term  and  the  ratio* 
and  equation  (B)  becomes 


By  change  of  signs       S  = 


r—l 

a 


This  gives  for  the  sum  of  a  decreasing  infinite  Bcries,  the 
following 

Rule.     Divide  the  first  term  hy  the  difference   between 
unity  and  the  ratio. 

EXAMPLES   FOR   PRACTICE. 

1.  Find  the  value  of  1,  |,  /g,  &c.,  to  infinity. 

!NoTE.  a  =  1 ,  r  =  J. 

Arts.  4. 

2.  Find  the  exact  value  of  the  series,  2, 1,  ^,  &c.,  to  infinity. 

Ans.  4. 

3.  Find  the  exact  value  of  the  series,  6,  4,  <fec.,  to  infinity. 

Ans.  18. 

4.  Find  the  ejcact  value  of  the  decimal  .8333,  &c.,  to 
infinity. 

Note. — This  circulating  decimal  may  be  expressed  thus  :  -^^  _|_  ^g^  _(_ 
rTnT5+'  ^^'  Hence,  a  =  Vtt*  and  r  =  ^^,  In  all  cases  of  this  kind, 
the  repetendf  taken  with  its  local  value,  will  be  the  first  term ;  and  tht 
order  of  its  lowest  figure  will  indicate  the  fractional  ratio. 

Ans.  f 


270  GEOMETRICAL   PROGRESSION. 

6.  Find  the  value  of  .323232,  &c.,  to  infinity. 

«  ==  i%%f  ^^  =  To'o^oo  ;  therefore,  r  =  jl^.      Am.  ||. 

6.  Find  the  value  of  .777,  <fec.,  to  infinity.  Ans,  J. 

7.  Find  the  sum  of  the  infinite  series,  1  +  -  -j — ,  -\ — ,4-, 

x^       x^      af^ 

&c.  ,  x' 

Ans,  —, :,, 

jr  —  i 

8.  Find  the  sum   of  the  infinite  geometrical  progression, 

b^      b^      6*  b 

a  —  b  -\ i  H — ^  — ,  &e.,  in  which  the  ratio  is . 

a       a^      or  a 

a" 
Ans, 


a  +  b' 
GEOMETRICAL  MEAJS'S. 

S46*  If  we  take  any  three  terms  in  geometrical  progres- 
sion, as 

a,  ar,  ar\ 
wo  have 

a  X  ar'^  a^r^,  product  of  the  extremes ; 
{arf  ==  aV,  square  of  the  mean. 

If  we  take  four  terms, 

a,  ar^  ar^^  ar^^ 
we  have 

a  X  aif^  ^  a^r^f  product  of  the  extremes ; 
ar  X  ar^  ^  aV,  product  of  the  means. 

Hence, 

I.  When  three  numbers  are  in  geometrical  progression^ 
the  product  of  the  extremes  is  equal  to  the  square  of  the 
mean. 

II  Wlien  four  numbers  are  in  geometrical  progression j 
the  product  of  the  extremes  is  equal  to  the  product  of  the 
means. 

Note. — This  last  property  belonc;s  also  to  geometi-ical  proportion  But 
the  pupil  must  not  confoxxjiii  proportion  with  series. 


GEOMETRICAL  MEANS.  271 

a  :  ar  \i  b     :  br,    are  quantities  in  geometrical  proportion, 
a  :  ar  :    ar^ ;  ar^,  are  quantities  in  geometrical  progression, 

247.  Tc  find  the  geometrical  mean  of  two  numbers,  we 
bave  from  the  principle  enunciated  above  (I),  the  following 

"RuL^.     Multiply  the  extremes  together,  and  take  the  square 
root  of  the  product. 


EXAMPLES   FOR   PRACTICE. 

1.  Find  the  geometrical  mean  between  2  and  8. 


Ans.  v/2  X  8  =  4. 

2.  Find  the  geometrical  mean  between  8  and  12. 

Ans.  6. 

3.  Find  the  geometrical  mean  between  5  and  80. 

Ans.  20. 

4.  Find  the  geometrical  mean  between  a  and  6. 

Ans.   (aby, 

5.  Find  the  geometrical  mean  between  ^  and  9.    Ans.  |. 

6.  Find  the  geometrical  mean  between  da  and  27a. 

Ans.  9a, 

7.  Find  the  geometrical  mean  between  1  and  9.     Ans.  3. 

8.  Find  the  geometncal  mean  between  2  and  3. 

Ans.  v'S. 

9.  Find  the  geometrical  mean  betweenV'a^^  and  \^ax\ 

Ans.  >/  ax^ 


272  GEOMETRICAL   PROGRESSION. 


APPLICATIONS. 


948.  In  a  geometrical  progression  there  are  five  parts,  as 
follows : 

1st.   The  first  term  ; 

2d.    The  ratio  ; 

3(1.    The  number  of  terras  ; 

4th.  The  last  term  ; 

5th.  The  sum  of  the  terms ; 

Since  the  independent  equations, 


symbol 

a. 

a 

r. 

a 

n. 

ti 

L. 

ii 

B. 

L=ar^-' 

(A) 

S--L-a 

(B) 

contain  all  of  the  parts,  any  two  may  be  determined  when  the 
other  three  are  known. 

Note. — Since  n  enters  into  tbe  above  formulas  only  as  an  exponenty  the 
process  of  determining  it  requires  a  knowledge  of  logarithms,  and  must 
be  omitted  here. 

1.  The  sum  of  a  geometrical  progression  is  468,  the  number 
of  terms  is  4,  and  the  ratio  5  ;  what  is  the  first  term  ? 


Analysis.  From  the  example,  we 
have  468,  4,  and  5,  for  the  values  of 
S,  ?2,  and  r,  respectively  ;  ii  —  1  is  3, 
and  7'""^  is,  therefore,  equal  to  125, 
which  substituted  in  formula  (A),  *:^ives 
(1).  The  value  of  r,  substituted  in 
formula  (B),  gives  (2).  To  eliminate  L 
we  substitute  its  value,  125a,  in  i2\ 
and  obtain  (3),  which  reduced,  gives 
a  =  3. 


2.  The  sum  of  a  geometrical  series  is  847,  the  ratio  3,  and 
the  number  of  terms  5  ;  what  is  the  first  term  ?        Ans.  7. 


OPERATION. 

S  =  468 

n  =  4 

r=5 

L  =  125a 

0) 

408  ^^Y" 

(2) 

^gg^625«-a 
4 

(3) 

156a  =  468 

(4) 

a  =  3 

(5) 

PROBLEMS.  273 

3.  The  sum  of  a  geometrical  progression  is  6220,  the  ratio 
6,  and  the  number  of  terms  5  ;  what  is  the  last  term  ? 

Ans,  6184. 

4.  The  first  and  last  terms  of  a  geometrical  scries  are  2  and 
162,  and  the  number  of  terms  5  ;  required  the  ratio. 

1 

/L\«-i 
Note. — From  formula  (A)  we  nave  r^^l-~) 

Ans.   3. 

5.  The  first  term  of  a  geometrical  series  is  28,  the  last  term 
17500,  and  the  number  of  terms  6  ;  what  is  the  ratio  ? 

Ans.  6. 

6.  The  first  term  of  a  geometrical  series  is  32,  the  last  term 
4000,  and  the  number  of  terms  4  ;  what  is  the  ratio  ? 

Ans.  5. 

7.  Find  two  geometrical  means  between  4  and  256. 

Note. — Two  means  will  require /owr  terms.    Hence  w  =4,  from  which 
data  r  may  be  found,  as  above. 

Ans.  16  and  64. 

8.  Find  three  geometrical  means  between  1  and  16. 

Ans.  2,  4,  and  8. 


PROBLEMS  IN  GEOMETRICAL  PROGRESSION 

TO    WHICH    THE    FORMUIAS    (A)     AND     (B)     DO    NOT    DIRECTLY 

APPLY. 

249,  The  general  representation  of  the  terms  of  a  geome- 
trical progression  is  made  thus  : 

^f  ^y,  ^y\  ^¥\  &c-» 

in  which  x  is  the  first  term,  and  y  the  ratio.  But  when  par- 
ticular relations  are  given,  it  may  become  necessary  to  adopt 
a  different  notation. 


274  GEOMETRICAL  PKOGRESSION. 

When  three  terms  are  considered  in  the  problem,  they  may 
DC  represented  thus : 

Or,         x\      xy,  y\ 

Since,  in  each  case,  the  product  of  the  extremes  is  equal  to 
the  square  of  the  mean  (5S4<J,  I). 

When  there  are  four  terms,  we  may  adopt  the  following 
notation : . 

x^  y^ 

y'  "^^  ^'  x' 

for  the  product  of  the  extremes  is  equal  to  the  product  of  the 
means  (246,  II). 

1.  The  sum  of  three  numbers  in  geometrical  progression  is 
T,  and  the  sum  of  their  squares  is  21 ;  what  are  the  numbers  ? 

SOLUTION. 

First  condition,  x  -f  '^xy  +  y  =  7  (i) 

Second  condition,  rr^  +      xy  -{-  y^  =  21  (2) 

From  (1),                                   ^  +  y  =7  —  \^xy  (3) 

From  (2),                                  x'-f-  2/'  =  21  —  xy  (4) 

Squaring  (3),  x^  +  2xy  +  y"^  =  49  —  14 \/^  +  xy  (5) 

Taking  (4)  from   (5),  2xy  =  28  —  \Wxy  -f  2xy  (6) 

Reducing  (6),                              \/ xy  =  2,  the  mean,  (T) 

From  (T),                                         Zxy  =  12  (S) 

Taking  (s)  from  (2),    y^  —  2xy  +  ^^  =  9  (o) 

Extracting  square  root  of  (9),    y  —  x=  S  (lo; 

From  (3)  and  (7),                       y  -\-  x  =  d  (ii; 

Prom  (10)  and  (H)                             x  =1,  first  term,  (i2) 

A.nd,                                                 y  =  ^i  third  term,  (is) 
Note. — In  equation  (9),  y^  is  put  first  because  y  is  greater  than  x. 


PROBLEMS.  .     275 

2.  The  product  of  three  numbers  in  geometrical  progression 
is  64,  and  the  sum  of  their  cubes  is  584  ;  what  are  the  numbers  ? 


SOLUTION. 

Let 

^■^  ^Vi  y^      ^>G  the  numbers, 

First  condition, 

x'y'  =  64 

0) 

Second  condition, 

^-f 

r^^y^  4-  y^  ==  584 

(2i 

Adding  (i)  to  (2), 

x-«-f 

2.ry  +  /  =  648  (  =  324-2) 

(3) 

Extracting  the  square 

root, 

^  +  y»  =  18  n/2 

(4) 

Taking  3  times  (i)  from  (2),  .r^— ■ 

.2.r^?/-H  f  =  392  (=  196  2) 

(6) 

Square  root, 

y^  — a:^=14v/2 

(6) 

Equation  (4), 

3/S  4-  ^  =  18n/2 

(7) 

From  (6)  and  (7), 

x^  =  2%/2 

(8) 

Squaring  (8), 

a^==% 

(9) 

Cube  root, 

a;'  =  2,  1st  term, 

m 

From  (6)  and  (7), 

t/3  =  16v/2 

(11) 

Squaring  (ii), 

2/^=512 

(12) 

Cube  root, 

t/^  =  8,  3d  term, 

(13) 

From  (10)  and  a3), 

a-y  ==  16 

(14) 

Square  root, 

xy  ~  4,  2d  term. 

(15) 

3.  Of  three  numbers  in  geometrical  progression,  the  sum 
of  the  first  and  second  is  90,  and  the  sum  of  the  second  and 
third  is  180  ;  what  are  the  numbers  ? 

Note.     Represent  the  numbers  by  x,  a:y,  and  ary*. 

Ans.  30,  60,  and  120. 

4.  The  sum  of  the  first  and  third  of  four  numbers  in  geo- 
metrical progression  is  20,  and  the  sum  of  the  second  and 
fourth  is  60  ;  what  are  the  numbers  ?        Ans.  2,  6,  18,  54. 

5.  Divide  the  number  210  into  three  parts,  such  that  the  last 
shall  exceed  the  first  by  90,  and  the  parts  be  in  geometrical 
progression.  Ans.  SO,  60,  and  12Q 


276  GEOMETRICAL   PROGRESSION. 

6.  The  sum  of  four  numbers  iu  geometrical  progression  in 
80,  and  the  last  term  divided  by  the  sum  of  the  mean  terms 
is  1^  ;   what  are  the  numbers  ?  Ans.  2,  4,  8,  and  16. 

7.  The  sum  of  the  first  and  third  of  four  numbers  in  geo- 
metrical progression  is  148,  and  the  sum  of  the  second  and 
fourth  is  888  ;  what  are  the  numbers  ? 

Ans.  4,  24,  144,  and  864 

8.  The  continued  product  of  three  numbers  in  geometrical 
progression  is  216,  and  the  sura  of  the  squares  of  the  extremes 
is  328  ;  what  are  the  numbers  ?  Ans.  2,  6,  18. 

9.  The  sum  of  three  numbers  in  geometrical  progression  is 
13,  and  the  sum  of  the  extremes  being  multiplied  by  the  mean, 
the  product  is  30  ;  what  are  the  numbers  ? 

Ans.  1,  3,  and  9. 

10.  Of  three  numbers  in  geometrical  progression,  the  sum 
of  the  first  and  last  is  52,  and  the  square  of  the  mean  is 
100;  what  are  th-e  numbers?  Ans,  2,  10,  50. 

11.  There  are  three  numbers  in  geometrical  progression; 
their  sum  is  31,  and  the  sum  of  the  squares  of  the  first  and 
last  is  626  ;  what  are  the  numbers  ?  Ans.  1,  5,  25. 

1 2.  It  is  required  to  find  three  numbers  in  geometrical  pro- 
gression, such  that  their  sum  shall  be  14,  and  the  sum  of  their 
squares  84.  Aris.  2,  4,  and  8. 

13.  Of  four  numbers  in  geometrical  progression,  the  second 
number  is  less  than  the  fourth  by  24,  and  tlie  sum  of  the  ex- 
tremes is  to  the  sum  of  the  means,  as  7  to  3  ;  what  are  the 
numbers  ?  Ans.  1,  3,  9,  and  27. 

14.  The  sum  of  four  numbers  in  geometrical  progression  is 
equal  to  their  common  ratio  +  1,  and  the  first  term  is  y'^  ; 
what  are  the  numbers?  Ans.  -j^^,  j^^,  y^^,  f  J. 

15.  How  much  will  $500  amount  to  in  4  years,  at  7  per 

cent,  compound  interest  ? 

Note. — Since  we  have  the  principal  at  the  commencement,  and  the 
trst  year's  amount  at  the  end  of  the  first  year,  the  number  of  terras  is  5. 
The  ratio  is  ^1.07,  equal  to  the  amount  of  $1.00  for  one  year. 

Ans.  $655,398 


PROPORTION.  277 


PROPORTION. 

2o0.  Two  quantities  of  the  same  kind  may  be  compared, 
and  their  numerical  rehation  determined,  by  seeking  how  many 
times  one  contains  the  other.  The  rehition  of  four  quantities 
may  be  determined  by  comparing  the  relation  of  two  of  like 
kind,  with  tv^o  others  of  like  kind.  These  comparisons  give 
rise  to  ratio  and  proportion. 

2^1.  Ratio  is  the  quotient  of  one  quantity  divided  by  an- 
other of  the  same  kind,  regarded  as  the  standard  of  comparison. 

There  are  two  methods  of  indicating  ratio. 

1st.  By  TVTiting  the  divisor  and  dividend  with  two  dots  be- 
tween them ;  thus, 

a  :  b 

is  the  indicated  ratio  of  a  to  6,  in  which  a  is  the  divisor,  and  b 
the  dividend. 

2d.  In  the  form  of  a  fraction  ;  thus,  the  above  ratio  becomes 

a 

3f>S.  Proportion  is  an  equality  of  ratios ;  thus,  if  two 
quantities,  a  and  b,  have  the  same  ratio  as  two  other  quantities, 
c  and  d,  the  four  quantities,  a,  b,  c,  and  d,  are  said  to  lye  pro- 
portional. 

Proportion  is  written  in  two  ways  ;  thus, 

a  :  b  :  :  c  :  df 

which  is  read,  a  is  to  6,  as  c  is  to  cZ ;  or  thus, 
a  :b  ~  c  :  d, 

which  is  read,  the  ratio  of  a  to  6  is  equal  to  the  ratio  of  c  to  d. 

Note. — The  second  is  the  modern  method,  and  more  fitly  expresses  the 
nature  of  proportion. 

24  ■» 


278  PROPORTIOIS. 

fSSS»  A  Couplet  is  the  two  quantities  which  form  a  ratio. 

S34,  Tlie  Terms  of  a  proportion  are  the  four  quantities 
which  are  compared. 

25^.  The  Antecedents  in  a  proportion  are  the  first  terms 
of  the  two  couplets  ;  or  the  first  and  third  terms  of  the  pro- 
portion. 

S^o6.  The  Consequents  in  a  proportion  are  the  second  term.. 
of  the  two  couplets ;  or  the  second  and  fourth  terms  of  tlie 
proportion. 

^^7.  The  Extremes  in  a  proportion  are  the  first  and 
fourth  terms. 

S58.  The  Means  in  a  proportion  are  the  second  and  third 
terms. 

Sof^.  A  Mean  Proportional  between  two  quantities  is  a 
quantity  to  which  tlie  first  of  the  two  given  quantities  has  the 
same  ratio  as  the  quantity  itself  has  to  the  second ;  thus,  if 

a  :  6  =  6  :  e 

the  quantity,  6,  is  a  mean  proportional  between  a  and  c ;  and 
the  three  quantities  are  said  to  be  in  continued  pro2:)ortion 

^HO.  A  Proposition  is  the  statement  of  a  truth  to  be  de- 
monstrated,  or  of  an  operation  to  be  performed. 

361.  A  Scholium  is  any  remark  showing  the  application 
or  limitation  of  a  preceding  proposition. 

3©S.  If  in  the  proportion 

a  :  h  =  c  :  d 

tie   2d  method  of  indicating  ratio  be  employed,  we  shall  ha"ve 

b      d 

-  =  -       (A) 

a      c 


GENERAL  PRINCIPLES  OF  PROPORTION.  279 

which  is  the  fundamental  equation  of  proportion.  And  any 
proposition  relating;  to  proportion  will  be  proved^  when  shown 
to  be  consistent  with  this  equation. 


GENERAL  PRINCIPLES  OF  PROPORTION. 
PROPOSITION   I. 

^6S«  In  every  proportion,  the  product  of  the  extremes  is 
equal  to  the  product  of  the  means. 

Let  a:h  =^  cidj  represent  any  proportion  ; 

Then  by  formula  (a),         —  =  -  ; 
■  a       c 

Clearing  of  fractions,      he  =  ad. 

That  is,  the  product  of  h  and  c,  the  means,  is  equal  to  the 
product  of  a  and  c?,  the  extremes. 

SCHOLIUM. — From  the  last  equation,  we  have 

ad  ^ 


The  first  mean,  h  = 

G 

The  second  mean,         ^  ~  "T 

he' 
d  I 

ha 


The  first  extreme,         a  —    •, 

a 


(1). 


(2). 


The  second  extreme,     d  = 

a 


Hence, 


1st.  Either  mean  is  equal  to  the  product  of  the  extremes 
divided  hy  the  other  mean.  (i). 

2d.  Either  extreme  is  equal  to  the  product  of  the  means 
divided  hy  the  other  extreme.  (2). 


280  PROPORTION. 


PROPOSITION  n. 

2S4.  Conversely.  If  the  product  of  hvo  quantities  be 
equal  to  the  product  of  two  others,  then  two  of  them  may  he 
taken  for  the  means,  and  the  other  two  for  the  extremes  of  a 
p)roportion. 

Let  be  =  ad         (i) 


Dividing  by  c,  6  =  —  (2) 


b 

_ad 
c 

b 
a 

_d 

~  c 

a 

:6  = 

Dividing  (2)  by  a,  -  =  -  (3) 

Hence  by  formula  (A),     a  :  b  =  c  :  d 

in  which  the  factors  of  the  first  product,  cb,  are  the  means, 
and  the  factors  of  the  second  product,  ad,  are  the  extremes. 

PROPOSITION   III. 

@llo.  If  four  quantities  he  in  proportion,  they  will  be  in 
proportion  by  alternation  ;  that  is,  the  antecedents  will  he  to 
each  other  as  the  consequents. 

Let  a  :  b  —  c  :  d 

Then  by  formula  (a),      -  =  -  (i) 

a       c 

be 
Multiplying  (D  by  c,     —  =  d!  (2) 

Dividing  (2)  by  5,  =  -  (3) 

Hence,  a  :  c  =  b  :  d 

in  which  a  and  c,  the  antecedents  of  the  given  proportion,  are 
proportional  to  b  and  d,  the  consequents  of  the  given  pro- 
portion. 


GENERAL  PRINCIPLES  OF  PROPORTION.  281 

Scholium. — A  proportion  and  an  equation  may  be  regarded 
as  but  different  forms  for  the  same  expression,  and  every 
equation  may  be  converted  into  a  proportion  under  various 
forms.     For  example, 

Let,  xy  =  a{a  +  h) 

Then,      x  :  a  =  a  -}-  b  :  y 

Or,  xy  :  a  =  a  -^  b  :  I 
Or,  a  :  X  =  y  :  (a  -i-  b) 


PROPOSITION   IV. 

266.  1/  four  quantities  be  in  proportion,  they  will  be  in 
proportion  by  inversion  ;  that  is,  the  second  will  be  to  the  first, 
as  the  fourth  to  the  third. 

Let  a  :  b  =  c  :  d 

Then  by  formula  (a),  -  =  -  (i) 

•^  a       c 

Clearing  of  fractions,  be  =  ad         (2) 

Hence  by  Prop.  II,        b  :  a  :  :  d  :  c. 


PROPOSITION    V. 

307.  If  three  quantities  be  in  continued  proportion,  the 
product  of  the  extremes  is  equal  to  the  square  of  the  mean. 

Let,  a  :  b  —    b  :  c 

By  Prop.  I,      ac  =  hb  =  ¥ 

Scholium. — Extracting  the  square  root  of  the  last  equation, 
vre  have 

b  =  V ac\  hence. 

The  mean  prop)ortional  between  two  quantities  is  equal  to 
the  square  root  of  their  product 
24* 


282  PROPORTION. 

PROPOSITION   VI. 

268.   Quantities    which   are    loroportional   to   the  same 
quantities  are  j^rojooriional  to  each  other. 


If 

a 

:  5  =  P  : 

:Q 

(A) 

And 

c 

:d=P 

:Q 

(B) 

We  are  to  prove 

that 

a 

:b=c  ; 

id 

From  (A) 

Form  (B) 

d      Q 

Hence,  (Ax.  7), 

b  _d 
a  '^  c 

Or, 

a 

:  b  =  c  : 

d 

PROPOSITION   VII. 

269.  If  four  magnitudes  be  in  proportion,  they  must  be  in 
proportion  by  composition  or  division  ;  that  is,  the  first  is  to 
the  sum  of  the  first  and  second,  as  the  third  is  to  the  sum  of 
the  third  and  fourth  ;  or,  the  first  is  to  the  difference  between 
the  first  and  second,  as  the  third  is  to  the  difference  between 
the  third  and  fourth. 

If  a  lb  =^  c  :  d  (A) 

We  are  to  prove  that   aia±b=^G\cdcid 


From  (A), 

b  _d 

a"'  c 

(1) 

Take 

1  ==  1 

C2) 

Adding  (i)  to  (2), 

i  +  Ui  +  1 

a                c 

(8> 

Subtractmg  (i)  from 

(2), 

a                c 

(4) 

From  (3), 

a  +  b        c  -h  d 
a                c 

(6) 

From  (4), 

a  —  b       c  —  d 
a                c 

(«) 

Hence,  from  (5), 

a 

I  a  ■■{-  b  =  c  :  G  +  d 

And  from  (6), 

a 

I  a  —  b  =  c  :  c  —  d. 

GENERAL   PRINCIPLES   OF   PROPORTION.  283 

Scholium. — This  composition  may  be  carried  to  almost  any 

extent,  as  we  see  by  the  following  iiivestigatiou ; 


Take  the  equation, 

b   _d 

a         c 

a) 

Multiplying 

by 

m  =  VI 

And 

vib       md 
a    ""    c 

(2) 

Adding, 

* 

n  =  n 

And 

n 

-f 

m6             .    md 

—  =  n  H 

a                    c 

(8) 

Reducing, 

71  a 

a 

mb       nc  -}-  md 
c 

(4) 

Hence, 

a  :  7ia 

-f  mb  =  c  :  nc  -f  md 

Or,  subtracting  (2)  from  n  =  w,  and  proceeding  as  before, 
we  shall  have 

a  :  na  —  mb  =  c  ;  nc  —  md 


PROPOSITION    VIII. 

S70.  If  four  quantities  be  in  proportion,  the  sum  of  the 
two  quantities  which  form  the  first  coujjlet  is  to  their  dif- 
ference, as  the  sum  of  the  tivo  quantities  which  form  the 
second  couplet  is  to  their  difference. 

If  a  :  b  =  c  :  d  (A) 

We  are  to  prove  that     a  -{-  b  :  a  —  6  =  c-f(/:c  —  d 

From  (A),  by  Prop.  YII,        a  :  a  +  b  =  c  :  c  -j~  d  (i) 

Also,  a  :  a  —  6  =  c  :  c  —  d  (2) 

From  0),  — ■ —  =  — —  (3) 

a  c 

^  a  —  be  —  d 

From  (2)  =  (4) 

a  c 

Tx.  .  T  ,.  ^  —  be  —  d 

Dividmg  (4),  by  (S),  — — -  =  — — -  (51 

a  -j-  0       c  -j-  a 

Whence  a  +  b  :  a  —  b  =  c  +  d  :  c  —  d 


284  PROPORTION 


PROPOSITION  IX. 

!I71.  If  four  quantities  he  in  proportion,  either  couplet 
may  he  muliiplied  or  divided  hy  any  number  whatever,  and 
the  quantities  will  still  he  in  proportion 

Let  a  :  b  =  c  :  d 

h        d 

Then,  -  =  -     (i;. 

a       c 

Multiplying  both  numerator  and  denominator  of  either  of 
these  fractions  by  any  number,  n,  (105,  III), 

We  have, 

Also, 

Hence  from  (-2),       na  :  nb  =  c  :  d 

And  from  {^\  a  :  b  =  nc  :  nd] 

in  which,  if  n  represent  a  whole  number,  the  couplets  are  mul- 
tiplied; and  if  71  represent  a  fraction,  the  couplets  are  divided. 


PROPOSITION   X. 

S7S.  If  four  quantities  be  in  proportion,  either  the  ante- 
cedents or  the  consequents  may  be  multiplied  by  any  number 
and  the  four  quantities  will  still  be  in  pi^oportion. 

Let  a:b  =c  id 

Then,  b        d 

Take 

Multiplying  (i)  by  (2),  (105,  I), 


nh 

d 

= 

(2) 

na 

G 

h 

nd 

— 

= 

:    

(3) 

a 

no 

Dividing  :i)  by  (2),  (lOti,  II), 

Hence  from  (3\ 
And  from  (4), 


a 

= 

c 

(1) 

m 

= 

m 

(2) 

mb 

md 

), 

— 

—— 

(3) 

/' 

a 

c 

h 

d 

- — - 

:  

(4) 

"ilia 

mc 

a 

:  mb 

c :  md 

ma  :  b 

= 

mc : 

d 

GENERAL  PRINCIPLES  OF  PAOPORTION.  285 


PROPOSITION   XI. 

^7*1.  IJ  four  quantities  he  in  proportion j  like  powers  or 
roots  of  the  same  quantities  will  he  in  proportion. 

Let  a  ih  =^  c  :  d 

Then,  -=    -        (i) 

'  a       0        ^ ' 

nih  power  of  d),  ^=  ^»      (2) 

1  I 

5»       d^ 
nth  root  of  a),  —=  ~       {^ 

a«       en 
Hence,  from  (2),        a"  :  6"  =  c"  :  d"" 

And  from  (3),  a"  :  6"  =  c^  :  (Z« 

PROPOSITION   XIL 

fSl^^:.  If  four  quantities  in  proportion  he  multiplied  or 
divided  term  hy  term  hy  four  others  also  in  proportion^  the 
2:)roduct  or  quotient  will  still  form  a  proportion. 


If 

a  I  h  =  G  :  d 

rA) 

And 

X  :  y  =  m  :  n 

(B) 

We  are  to  prove  that 

ax  :  hy  =  cm  :  dn 

And 

a     b         G     d 

X  '  y  ~~  m  '  n 

From  (A), 

ad—  ho 

a) 

From  (B), 

xn  =  ym 

(2) 

Multiplying  (i)  by  (2), 

{ax')  (dn)  =  (h^J)  (cm) 

(3) 

Dividing  (i)  by  (2), 

\x  J  \nJ        \y  /  \m) 

(4) 

From  (3)  by  Prop.  II, 

ax  :  by  =?=  cm  :  dn 

And  from  (4), 

a     b        G     d 
X  '  y       m  '  n 

236  PROPORTION. 

PROPOSITION   XIII. 

275.  If  any  number  of  proportionals  have  the  same 
ratiOj  any  one  of  the  antecedents  will  be  to  its  consequent  as 
the  sum  of  all  the  antecedents  is  to  the  sum  of  all  the  conse- 
quents. 

Let  a  :  b  =  a  :  b         (A) 

iVlso,  a  :  b=  G  :  d         (B) 

a  :  b  =ni  :  n         [fi) 
&c.  =  &c. 
We  are  to  prove  that    a  :  b  ==  {a  -r  c  -}-  m)  i  (b  -{-  d '^-  ri) 

From  (A),  ab  =  ab 

From  (B),  ad  =  cb 

From  (C),  an^=  mb 


By  addition,     a(b  -f  c?  -f-  n)  =  b(a  -f  c  -j-  in) 

By  Prop.  II,  a  :b=  (a  +  G-i-m')  :  (b  +  d  +  n) 


PROBLEMS  IN  PROPORTION. 

2711.  1.  Find  two  numbers,  the  greater  of  which  is  to  the 
less  as  their  sum  to  42,  and  the  greater  to  the  less  as  their 
difference  is  to  6. 

SOLUTION. 

Let  X  =  greater ;  y  == 


By  the  conditions,  ] 


=  X  +  y  :  42  (1) 

X  —  y  :  Q  (2) 

Prop.  YI,                   X  +  y  :  4:2  =  X  —  y  :  6  (3) 

Prop.  Ill,            X  +  y  :  X  —  y  =  42  :  Q  (4) 

Prop.  YIII,                       2x:2y  ==  48  :  36  (5) 

Prop.  IX,                            X  :    y  =  4  :  S  (6) 

From  (1),  by  Prop.YI,           4  :  S  =  x  -}-  y  :  42  a) 

From  (2),  by  Prop.  YI,           4  :  '6  =  x  —  ^  :  6  (8) 

From  (7),  by  Prop.  I,         cc  4-  y  ==  56  (9) 

From  (8),  by  Prop.  I,         x  —  y  =  8  (lo) 
Hence,                                        cc  =  32 
And                                            y  =  24 


PROBLEMS. 


287 


2.  Divide  the  number  14  into  two  such  parts,  that  the  quo- 
tient of  the  greater  divided  by  the  less,  shall  be  to  the  less 
divided  by  the  greater,  as  100  to  16. 


Let 


SOLUTION. 

X  =  the  greater ; 
y  =  the  less 


By  the  conditions, 

From'(i),  by  Prop.  IX, 
Prop.  XI,' 
Hence,  from  (4), 
But 

Therefore, 


a;  =  10 
2/  =  4 


3.  Find  three  numbers,  in  geometrical  progression,  whose 
sum  is  13,  and  the  sum  of  the  extremes  is  to  the  double  of  the 
mean  as  10  to  6. 


Let 


SOLUTION. 

x,  xy,  xy'^^  represent  the  numbers. 


^                   ^.,.                             (   x    ■\-  xy -{■  xri"^—  13 
By  tne  conditions,                   \               v         j 

ix^f  +    .T  :  2:rt/  =  10  :  6 

(1) 

(2) 

From  (2),  by  Prop.  IX,                y^  -}-    I  :     2y  =  10  :  6 

(S) 

From  (3),byPr.TIII,  (2/^+22/4-1):  (2/'— 2y+l)x=  16  :  4 

4) 

Prop.  XI,                                  t/  +  l:2/  —  1  =  4:2 

(6) 

And                                                          21/   :   2  =  6    :  2 

(6) 

Hence,                                                              y  ==  3 

And                                                                  X  ==  1 

288  PROPORTION. 

4.  The  product  of  two  numbers  is  35,  and  the  difference  of 
their  cubes  is  to  the  cube  of  their  difference  as  109  to  4 ;  what 
are  the  numbers  ? 


SOLUTION. 

Let 

X   = 

:the 

greater,  and  y  =  the  less. 

By  tne  conditions, 

{.- 

f  : 

xy  = 

35 

109: 

:4 

(1) 

(2) 

Dividing  1st  couplet ) 
of  (2)  hj  x  —  y,          ) 

»'  +  ^2/  +  y' 

:(^  — ^//  = 

109; 

:4 

(3) 

Or,                       x'  +  xy  +  y^'. 

ix'^ 

-  2xy  +  2/^  = 

109; 

:4 

(4) 

From  (4)  Prop.  VII, 

Sxy 

:x'- 

^2xy  +  y'  = 

105 

:4 

(5) 

But,  (1) 

oxy  = 

105 

Hence,  from  (5) 

(x_yy  = 

4 

(6) 

And 

X  —  y  = 

,  2 

(7) 

From  (1)  and  (7), 

a:  +  2/  = 

12 

(8) 

Hence, 

x  = 

:7 

And 

2/  = 

5 

5.  What  two  numbers  are  those,  whose  difference  is  to  their 
sum  as  2  to  9,  and  whose  sum  is  to  their  product  as  18  to  77  ? 

Arts.  11  and  7. 

6.  Two  numbers  have  such  a  relation  to  each  other,  that  if 
4  be  added  to  each,  the  sums  will  be  to  each  other  as  3  to  4  ; 
and  if  4  be  subtracted  from  each,  the  remainders  will  be  to 
each  other  as  1  to  4  ;  what  are  the  numbers  ? 

Ans.  5  and  8. 

7.  Divide  the  number  16  into  two  such  parts,  that  their  pro- 
duct shall  be  to  the  sum  of  their  squares  as  15  to  34. 

Ans,   10  and  6. 

8.  There  are  two  numbers  whose  product  is  320,  and  the 
difference  of  their  cubes  is  to. the  cube  of  their  difference  as 
61  is  to  1 ;  what  are  the  numbers  ? 

Ans.   20  and  16, 


APPROXIMATE   ROOTS.  289 

APPROXIMATE    ROOTS 

OF   HIGHER  DEGREES. 

2?  7.  A  root  of  any  number  is  evidently  one  of  the  equal 
factors  which  compose  the  number.  Thus,  the  square  root  of 
25  is  one  of  the  two  equal  factors,  5x5,  which  produce  25. 
The  fifth  root  of  32  is  one  of  the  five  equal  factors,  2x2x2 
X  2  X  2,  which  produce  32. 

If  a  number  is  not  composed  of  as  many  equal  factors  as 
there  are  units  in  the  index  of  the  required  root,  the  number 
is  a  surdy  and  the  required  root  can  only  be  obtained  approxi- 
mately. This  may  be  done,  as  we  have  seen  (195j,  by  ex- 
tending the  general  method  of  extracting  the  root,  to  decimal 
periods.  Another  method,  and  the  one  we  are  now  about  to 
consider,  is  to  decompose  the  number  into  factors  nearly  equal, 
and  average  the  result, 

$578.  If  the  number  a  be  composed  of  3  factors,  each  equal 
to  X,  then 

g —  ^  —  V  a. 

In  like  manner,  if  the  number  a  be  composed  of  three  fac- 
tors, Xf  2/,  z,  nearly  equal  to  each  other,  it  is  obvious  that 

1 =  v^a,  nearly.  l|p 

To  illustrate  this  principle  by  numeral  examples,  we  have 
2  X  3  X  4  =  24 

A   A                       2  4-3+4 
And  q =  ^ 

But  v^24         n=  2.88  +. 

That  is,  one  third  of  the  sum  of  the  three  unequal  factors, 
2,  3  and  4,  which  compose  24,  is  3  ;  while  the  cube  root  of  24 
is  2  88  -f ,  a  number  less  than  3  by  only  .12. 
25  T 


290 

APPROXIMATE  ROOTS. 

Again,  tako 

5  X  7  X  8  =  280 

And 

''■l'-'  =  6M  + 

But 

v/280  =  6.54  + 

Difference, 

.12  4-. 

Similarly,  we 

shall  find  that  if  uxyz  =  a, 

Then 

u+x+y+z 

Hence,  in  general  terms. 

If  a  number  be  decomposed  into  n  factors,  nearly  equal  U} 
each  other,  the  nth  part  of  their  sum  will  be  nearly  equal  to 
the  nth  root  of  the  number, 

^79.  If  a  number  does  not  consist  of  factors  nearly  equal, 
or  if  it  is  a  prime  number,  it  may  be  decomposed  into  approxi* 
mate  factors. 

Thus,  if  xyz  =   a, 

Then  x  =  — , 

yz 

in  which,  if  y  and  z  be  assumed,  x  may  be  computed, 
i  L  Find  the  cube  root  of  100. 


OPERATION. 

Let 

xyz  =  100 

Assume 

y  =  4 

And 

z  =  5 

Then 

X 

100       100 

-    yz   ~    2Q   - 

Adding  A^alues  of  x+y  +  z,       4+5+5  =  14 

Hence,  dividing  by  3,  VlOO  =  4.66  +  ,  1st  approx. 


} 


2/  = 

4.641 

2  = 

4.642 

X  = 

4.64176  + 

APPROXIMATE  ROOTS.  291 

As  the  root  songbt  is  greater  than  4.6,  and  less  than  4.7,  it 
must  evidently  be  some  number  between  these  two.  We  will 
therefore 

IS'ext,  assume  2/=    ^-^ 

And,  z=    4.7 

T,e„.  .  =  l^=i^=    4.62503  + 

*  yz        21.62 

Adding  the  new  values,  x  +  y  -\-  z  =  13.92503+ 

Hence,  dividing  by  3,  ^JFO  =    4.64167  + , 2d approx. 

Next,  assume 

And, 

Then,  dividing  as  before, 

Adding  the  new  values,  oc  +  y  +  z  =  13.92476  + 

Dividing  by  3,  VWO  =  4.64158 +,  3d  approx. 

which  is  correct  to  tire  last  decimal  place. 

2    Find  the  square  root  of  3. 

OPERATION. 

Let  xy  =  3 

Assume  ^  =  1.6 

Then,  a:  =  A  =  1.875  + 

l.o 

Adding,  x  +  y  =  3.475  + 

Dividing  by  2,  n/3  =  1.7375,  1st  approximation. 

Next,  assume  y  =  1.732 

Then,  x  =  T-|ri=  1.7321016  + 

Adding,  x  +  y=  3.4641016  + 

Dividing  by  2,  \/3  =  1.7320508  +  ,  2d  approx. 

which  is  correct  to  the  last  decimal  place. 

Hence,  to  obtain  the  approximate  7?th  root  of  a  number,  we 
have  the  following 

KuLE.     I.  Assume  n  —  1  factors  as  near  the  required  root 
as  may  be  found  by  inspection, 

II.  Divide  the  given  number  by  the  product  of  the  assumed 
factors,  and  the  quotient  will  be  the  remaining  factor. 


292  APPROXIMATE   ROOTS. 

III.  Divide  the  sum  of  the  n  factors  thus  obtained,  by  n,  and 
the  quotient  ivilt  be  the  first  approximation  to  the  required  root, 

lY.  Assume  a  second  set  of.n  —  1  factors,  and  proceed  as 

before  ;  and  thus  continue^  till  the  desired  approximation  is 

obtained. 

Note.  1.  By  inspecting  the  two  examples  given,  it  will  be  seen  that 
the  second  approximation  is  less  than  the  lirst,  and  the  third  less  than  tho 
second.  Hence,  every  approximation  is  greater  than  the  true  root,  and 
this  principle  will  govern  the  selection  of  the  factors  to  be  assumed  iu 
each  step. 

2.  Also,  by  inspecting  the  two  preceding  operations  it  will  be  seen  that 
each  approximation  after  the  first  obtains  one  or  more  correct  decimal 
figures  in  the  required  root. 

EXAMPLES  FOR  PRACTICE. 

3.  Kequired  the  4th  root  of  18. 

OPERATION. 

Let  uxyz  =  18 


{" 


u  =  2 
=  2 


Assume, 

'      =  2 

Then  by  (II),  z  =  2.25 

Adding,  u  +  x  +  y  +  z  =  8.25 

Dividing  by  4  (III),        v/l8  =  2.0625,  1st  approx. 

2^06 
Next,  assume  -^  ^  =  2.06 


.!/  =  2.05 
Then,  as  before,  z  =  2.069113  + 

Adding,         u-^-  x  +  y  +_z  =  8.239113  + 

Dividing  by  4,  v/l8  =  2.059778  +,  2d  approx. 

XoTR. — It  will  bo  observed,  in  the  above  example,  that  16,  a  number 
near  18,  is  a  perfect  4th  power,  whose  root  is  2.  Hence,  ti,  x  and  y  were 
assumed  each  as  2. 

4.  Required  the  cube  root  of  130. 

5.  Required  the  4th  root  of  260. 

6.  Required  the  4th  root  of  640. 

7.  Required  the  5th  root  of  2. 

8.  Required  the  5th  root  of  38. 


APPROXIMATE  ROOTS. 


293 


QSO.  When  tlie  numbers  are  large,  or  the  index  of  tlie  root 
is  higher  than  the  3d  or  4th,  the  following  table  will  be  of  use 
in  assuming  the  factors  for  the  1st  approximation.  It  will  be 
seen  that  the  first  column  contains  certain  numbers,  taken  at 
intervals  between  1  and  270C0  ;  the  second  column  contains 
the  squares  of  the  numbers  in  the  third  ;  the  third  column 
contains  the  cube  roots  of  the  numbers  in  the  first ;  the  fourth 
contains  the  6th  roots ;  and  the  fifth,  the  9th  roots. 


A 

aI 

Ai 

A* 

Ai 

1  

1  ... 

...     1    ... 

...  1.0000000  ... 

...  l.OOOOOC 

8  

4  ... 

...   2  ... 

...  1.4142136  ... 

...  1.259921 

27  

9  ... 

...   8  ... 

...  1.7320508  ... 

...  1.442250 

64  

.   16  ... 

...   4  ... 

...  2.0000000  ... 

...  1.587401 

125  

25  ... 

...   5  ... 

...  2.2360680  ... 

...  1.709976 

216  

.   36  ... 

...   6  ... 

...  2.4494897  ... 

...  1.817121 

843  

.   49  ... 

...   7  ... 

...  2.6457513  ... 

...  1.912933 

512  

.   64  ... 

...   8  ... 

....  2.8284271  ... 

...  2.000000 

729  

.   81  ... 

...   9  ... 

...  8.0000000  ... 

...  2.080084 

1000  

.  100  ... 

...  10  ... 

...  3.1622777  ... 

...  2.154435 

1331  

.  121  ... 

...  11  ... 

...  3.3166248  ... 

...  2.223980 

1728  

.  144  ... 

...  12  ... 

...  3.4641016  ... 

...  2.289428 

2197  

.  169  ... 

...  13  ... 

...  3.6055513  ... 

...  2.351335 

2744  

.  196  ... 

...  14  ... 

...  3.7416574  ... 

...  2  410142 

3375  

.  225  ... 

...  15  ... 

...  3.8729833  ... 

...  2.466212 

4096  

,  256  ... 

...  16  ... 

...  4.0000000  ... 

...  2.519842 

4913  

,  289  ... 

...  17  ... 

...  4.1231056  ... 

...  2.571282 

5832  

,  324  ... 

...  18  ... 

...  4.2426407  ... 

...  2.620741 

6859  

.  361  ... 

...  19  ... 

...  4.3588989  ... 

...  2.668402 

8000  

,  400  ... 

...  20  ... 

...  4.4721360  ... 

...  2.714418 

9261  

.  441  ... 

...  21  ... 

...  4.5825757  ... 

...  2.758923 

10648  

,  484  ... 

22 

...  4.6904158  ... 

...  2.802039 

12167  

.  529  ... 

...  23  ... 

...  4.7958315  ... 

...  2.843867 

13824  

,  676  ... 

...  24  ... 

...  4,8989795  ... 

...  2.884499 

15625  

625  ... 

...  25  ... 

...  5.0000000  ... 

...  2.924018 

17576  

.  676  ... 

...  26  ... 

...  5.0990195  ... 

...  2.962496 

19683  

729  ... 

...  27  ... 

...  5.1961524  ... 

...  3.000000 

21952  

784  ... 

...  28  ... 

...  5.2915026  ... 

...  3.036589 

24389  

841  ... 

...  29  ... 

...  5.3851648  ... 

...  3.072317 

27000  

900  ... 

...  30  ... 

...  5.4772256  ... 

...  8.107232 

25* 

294  APPROXIMATE   ROOTS. 

1.  Find  the  7tli  root  of  2211. 

Referring  this  number  to  the  table,  we  find  the  number 
nearest  to  it  in  column  A  is  2197,  whose  Gth  root  (A^),  is 
is  3.6  -f ,  and  whose  9th  root  (A^),  is  2.3  +.  Therefore,  the 
7th  root  of  2211  must  be  nearly  equal  to  3,  and  we  will  tako 
this  number  for  each  of  the  six  factors  to  be  assumed.     Thus, 

OPERATION. 

Let  3  X  3  X  3  X  3  X  3  X  3.2;  =  2211 

3)2211 
3)"737 

3)  245.666666 
3)  81.888888 
3)  27.296296 
3)       9.098765 

37032921  =  0?. 

Hence,         ^22n  =  3  +  3  +  3  +  3  +  iT"3T^:^2921 

7 
_  21.032921 

~"         7 

=  3.004703,  1st  approximation. 

For  the  second  approximation,  we  may  assume  3.004  for  each 
of  the  6  equal  factors.  As  the  method  above  used  would  be 
somewhat  tedious,  we  may  avoid  the  labor,  by  an  application 
of  the  binomial  theorem. 

For  this  purpose  let  it  be  observed  that  the  powers  of  a  frac- 
tion are  less  than  the  fraction  itself.     Thus, 

(.iy  =  .oi 

(.1)3  =  ,001 
(.1)*=.0001 

We  have  found  that  the  7th  root  of  2211  is  greater  than  3, 
by  a  small  fraction.  Let  this  fraction  be  represented  by  x ; 
end  we  shall  have  the  following  equation  : 

(3  +  xy  =  2211. 


APPROXIMATE    ROOTS.  295 

Expanding  the  first  member,  indicating  the  powers  of  3,  we 
have 

2187  +  7(3^^  +  21(3)^0:2  +  85(3)^:c^  +35(3yx*  -f- 
21{Syx'  +  7(3>c«  +  x'^  =  2211. 

Now  as  a:  is  a  small  fraction,  the  powers  of  x  will  be  still 
Bmaller.  And  if  the  terms  containing  x^j  and  all  the  higher 
powers  of  x  be  omitted,  the  equation  will  still  be  approxi- 
mately true  ;  and  we  shall  have 

2187  +  7(Syx  +  21(3)5x'^  ==  2211  a) 

Transposing,  7(3)«:c  +  21(3)V  =  24  (2) 

Dividing  (2)  by  3,      7(3/^  +  2l(3yx^  =  8  (3) 

Expanding  (3),  1701a:  +  1701a;''    =  8  (4) 

Dividhig  (4)  by  1701,  x'  +  x  =    .0047081  (5) 

Completing  the  square,  4x2+  ()  -f  1  =  1.0188124  (6) 

Extracting  square  root,  2x  +  1  =  1.00936  (7) 

Reducing  (7),  x  =    .00468 

Hence,  v^22Ti  ==  8  +  a:  =  3.00468,  2d  approx. 

2.  Find  the  7th  root  of  2412. 

Referring  to  the  table,  we  find  this  number  in  column  A, 
between  2197  and  2744;  and  by  examining  the  6th  and  9th 
roots  of  these  numbers,  we  find  that  the  7th  root  of  the  given 
number  must  be  a  little  greater,  or  a  little  less,  than  3. 
Hence, 

OPERATION. 

Let  3  -f  a;  =  required  root. 

Put  a  s=  3 

Then,  (a  +  xf  =  2412 
EKpanding,  a;+7a«a;+21alx'+35a*a-»  +  &c.  =  2412 

Or,  approximately,  a'  +  7a^x  =  2412           (i; 

Restoring  value  of  a  in  (i),  2187  +  5103x  =  2412           (2) 

Transposing  5103.C  =  225             (3) 

And  X  =.044  -f 

Hence,  S  +  x  =  3.044  + 


296  APPROXIMATE  ROOTS. 

That  is,  the  seventh  root  of  2412  is  very  nearly  3.044;  but 
this  is  a  little  too  large.  If  we  would  find  the  root  more  ac- 
curately we  will  place  3.04  for  a,  or  (3  +  m)  in  equation  (i). 
Then  that  equation  will  become 

(3  +  m)^  +  7  (3  +  my  x  =  2412. 

Taking  the  higher  terms  of  (3  +  m)',  and  (3  +  my  ip 
separate  columns,  we  have 


3^  =  2187 

8«  =  729 

7(3>i  =    204.12 

6(3)»m=    58.32 

21(3ym«  =        8.1648 

15(3)*m^  =      1.944 

85(3)*m»  =          .18144 

20(3)»m»=       .03456 

S5(3)»m*  =          .0024192 

15(3)''m*  =       .0003456 

789.2989056 

Adding,    2399.4686592  +  7(789.2989056)a;  =  2412 
Whence,  from  the  above  equation,  we  have  approximately 

2399.4686592  +  7(789.2989056)a;  =  2412 ; 
^  12.5313408  _  „^__  ^ 

Q^-^=7(m2989056)  =  '^^^^^^+> 
Hence,  ^2412  =  3.042268 +. 


EXAMPLES  FOR  PRACTICE. 

8.  What  is  the  cube  root  of  14000  ? 

4.  What  is  the  5th  root  of  812  ?  Am.  3.81893. 

5.  What  is  the  8th  root  of  1340  ? 

6.  What  is  the  7th  root  of  9150 1 


MISCELLANEOUS   EXAMPLES.  297 


MISCELLANEOUS   EXAMPLES. 


1    What  is  the  value  of —  ,  when  a  =  4, 

5  =  5,  and  m  =  20  ?         m^b--(l'—  m)a 

A71S,    -J. 

2.  What  is  the  sum  of  (a  +  b)s^ax  +  12(a  +  b)^^ax  -— 
7(a  -f  b)^ax  4n(a  -f  5)v^ax? 

^ns.   fQ(^a  +  &)  +  4n(a  +  6)")  >/a~x ; 
Or,  (6  +  4n)  (a  +  6)>/ax. 

8.  What  is  the  sum  of  (A  +  B)^  (x  +  y)  and  (A  —  B)*- 
(a;  +  2/)  ?  Ans.  2{A.'  +  B'O  (x  +  y\ 

4.  What  is  the  sum  of  ba'b  +  Sa-^fe^c,  6a*6  +  2a-262(?,  4- 
10  ab,  and  9a*6  —  8a-=^6^c  —  10a6  ?  ^ 

Ans,  20a'b—  •^. 

5.  From    v^x^  —  ^/^  +  4(0-  +  y)  —  Sy^a  +  a;,  subtract 
3(a;  -♦-  2/)  —  2(:r^  —  y'y  -h  13(a  +  a^)l       ' 

^ns.  Sv^ic^  —  2/^  +  03  +  2/  —  16v^a  +  ir» 

6.  From  17ax*  +  Say  +  10a,  subtract  7ax'^  +  4ay  +  12a 
—  2ba,  Ans,   (lOx^  _  y  -^  2  +  26)a. 

7.  What  is   the   value   of  (2x')  (j^2xy)  (—2mx)  (—2xy^) 
(2m*xy)  ?  Ans,  —  (2mxyy 

8.  Multiply  a»  —  a'6  +  afe^  —  ft^  13^  ^  ^  j 

-<4ns.  a*  —  b\ 
9    Multiply  as*"  --  a;**-^  ^as^'V  —  'f^hj  x  +  y. 


208  MISCELLANEOUS  EXAMPLES. 

10.  Expand  (m  —  z--  1)  (m  +  1)  (m'  +  1). 

Ans.  m*  —  m^z  —  m^z  —  mz  —  z  —  1. 

11.  Expand  (ox'^i/  —  oxy^)  (Sx^y  —  Sxy^), 

A71S.  ^xY  —  ISxy  +  9^y, 

12.  Expand  (c*"  +  c")  (c^"  +  c"). 

Ans,  c^"'  4-  2c'"+"  4-  cr^. 

13.  Find  two  factors  of  x^  +  3/*. 

^?i6*.   {x  +  y)  and  a;*  —  xhj  -{-  x^y^  —  xy^  +  y\ 

14.  Find  two  factors  of  x^  —  y^. 

Ans.   (xT-  +2/")  and  ix'  — 2/')  ; 
Or,  {x  —  y)  and  x^  -\-  x^y  4-  ct-?/'  +  a:'^/^  +  1/*. 

15.  Multiply  (a'  —  ah  +   a6^  —  Jh'  +  Jh'  —  h")  by 
(  a*  +  &).  Prod  a^  _  h\ 

16.  Find  the  factors  of  a^  —  h\ 

Ans,   (a'—b)  (a^+a'b  +  a'b'+b^),  or  (a'  +  b')  (a'—b'). 

17.  Find  the  greatest  common  factor  of  (a*  —  1),  (a^  4-  a^), 
(a«  4-  1).  Ans,  0}  +  1. 

18.  Find  the  greatest   common   divisor  of  a"  —  5ah  4-  46'' 
and  a^  —  ab  +  ^a¥  —  Sb^,  Ans,  a  —  b, 

19.  Find  the  sum  of  the  following  fractions;   indicating 
that  sum  by  S,  and  condensing  it  to  a  single  term. 

a^  h  ^        ab  ,         , 

-,  and   — -TT-o-  -^^s.   1. 


(a  +  6)"  a  +  b'  (a  +  b) 

20.  What  is  the  sum  of  the  following  fractions  ? 

_^^,  ^_,  and -i-.     Ans.^-^L±q^±l. 
x^  —  y^  X  '\-  y  X  —  y  x^  —  y^ 

21.  Divide  51a'(c  —  1)"*+"  by  ITa^c  — 1)"^-". 

Ans,  3(c  — 1)*~. 

22.  Divide  5(a  —  1)  4-  6(1  —  of  by  1  —  a. 

Ans,  1  —  6a. 


MISCELLANEOUS  EXAMPLES.  299 

23.  Divide  771  —  ?i  by  v'm  —  >/ n,       Ans.  v^wi  -1-  v^n. 

24.  Divide  a'  +  ab  +  b^  hj  a -{•  ^^a'b  +  b, 

Ans,  a  —  v^a6  +  b 

25.  Factor  a;^y--a:y'—-a:y  +  iP—y. 

26    What  is  the  greatest  common  divisor  of  ia^z — \2amz^+ 
%abcz  and  ba^nix —  Ibm'^xz  +  10 bcmx  ? 

Ans.  a^  —  Smz  +  26c. 

27.  What  is  the  least  common  multiple  of  Sd^x,  4:axy,  and 
16ax^  ?  Ans.  4:SaVy. 

28.  What  is  the  least  common  multiple  of  ac  —  om%  o?c  + 
Zacnv  and  a^c  —  9cm*  ?  Ans.  a^c  —  9ac7M*. 

s/aFb '^ai/ 

29.  Reduce  — ^r-, — ; — r.,  to  its  lowest  terms. 

a^  —  2ab  +  b^ 

^ab 
Ans.  --. 

_  ^  —  ^ 

Ttx(  ^^  CL  — —  ^^   b  "^ 

SO.  Reduce  — -^=z =t— ,-^  to  an  entire  form. 

Ans.  m(a  +  6)  —  2m  s/ob, 
81    Find  the  value  of  ^^  -  (^^  +  ^). 

Ans.  ^^. 
82.  From  i  +  i  +  |=-J,  subtract  \  +  ^^^. 

33.  Divide  1  -\ -— :r,  by  1 — ^.  u4ns.  n. 

n  +  1     "^  71+1 

J5^0TE. — The  product  of  the  divisor  and  quotient  must  be  equal  to  the 
dividend.     Let  Q  =  the  quotient,  then 


«c-j^l)-'+:-^- 


300  MISCELLANEOUS   EXi^MPLES. 

34.  Divide-^ ;>  by  (a; ; — ).  Ans. . 

or  —  x^        \        a  +  x/  a  —  x 


35.  Simplify  the  fraction,  "t!  "^  :-^.  Ans.  ^^^ 


tax 


a  —  x  a-^x 

86.  Divide  2 ,  by — :: — . 

87.  What  is  the  value  of  ^?L±^|:z;(^g!  y 

Ans. 


a  —  b' 


88.  What  is  the  value  of  ^^^IJt^tt^^I  ? 

(1  +  by  —  (1  —  by 


39.  Raise  v^a  +  6  to  the  4th  power.         Ans.  (a  +  6)'. 

40.  Raise  a  V  —  1  to  the  3d  power.    Ans.       d\/  —  1. 

41.  Raise  2a  +  36  to  the  9th  power. 

Ans.  h\1a^  +  256-27a'6  +  512-81a'6''  +  256-567a'6'  +  &c. 

Reduce  the  following  equations : 

42.  {x  +  Va)'  —  (ar  —  \/a)'  =  2a>/^. 


A                            «^^ 

43. 

m  +  2c 

44. 

6       v^a 6  -f  v'a 

a?         aj            x/a6> 

u4ns.  X  =  by^a. 

45. 

4           a:  +  1       ar  4-  1 

ar  — 1^      4       "^      4      • 

Ans.  a;  =  d=  3. 

46. 

-g — — rr  a=x  5a?  —  5.           An». 

aj  ==3  zfc  1.224744 -f. 

MISCELLANEOUS   EXAMPLES.  gOt 

47.  -5 =  =  _ZL_.  jins.  X  ==  ±*n/2  ±:2v/5 

48.  vsr=9  =  :;7=fg.  ^n«.  X  =  9. 

49.  (1  +  xy  +  (1  —  a?)''  =  242.  _ 

Ans.  ic  =  dh  2  or  it  v^ —  6. 

60.  Find  the  6th  power  of  \^2  -f  "^3,  or  expand  (v^2i  -f 
^3y)®,  and  afterward  suppress  all  the  powers  of  x  and  y,  on 
the  supposition  that  each  is  equal  to  unity. 

Ans,  8  +  24>/6-f  180  + 120\/6 4-270 +  54 v/6+  27. 

ifl 9 

51.  Given  - — -—^  =  7  —  v,  to  find  the  value  of  y, 

Ans,  2/  =  5. 

25-;p« 9 

b2.  Given  -= — -—^  =7  —  5x,  to  find  the  value  of  x. 
oa?  +  3 

u4ws.  X  =  1. 

63.  Given  |-+  2  =  |(^-  1  )  +  ^^^Y^f  to  find  the 
value  of  X.  Ans,  x  =  o. 

54.  Given  x»  +  2x»  +  cc  =  (a»  +  3a:)  (a;  —  1)  +  32,  to 
find  the  value  of  x.  Ans.  x  =z  S. 

65.  Given  (x  +  ly  —  (x  —  !)«  =  l344x. 

Ans.  a;  =  it  3,  or  d=  4/—  -^^ 

_^      (ox  +  ey  =  2aV  7  f  x  =  ac^. 

Ccx  +  ay  =s  ac^c'  +  a')>  Cy  ==  ^^« 

Tcc+y  —  2=»4^  rx==3. 

58.    <2x  +  2z  —  y==d>  Ans.    <  y  =^  2. 

I  3t/  +  32:  —  X  «  6  3  (  z  «  1. 
26 


^Q2  MISCELLANEOUS    EXAMPLES. 

lu  +  x+y=^^a  —  6m  \  I  u  =  a  '--    m. 

^,    \u  4-  X  +  z  =  Sa  —  1m {  ,        ]x  ^  a  —  2m, 

)u  +  y  +  z=oa  —  8m  (  )  2/  =  «  —  ^^^• 

\x  +  y  +  z  =  Sa  —  9m  /  [z  =  a  —  4  ;?i. 

^"-    l4a;»_42/  =  28i  ^"'-    1 2/ =  2,  or       6J-. 

f  ck'  -r  37/'  =  28  •>  f  a;  =  5,  or    v/2i. 

^"    l  a;' +  2ccv  =  35  i  ^"*-     U  =  1,  or  iv'21. 


62. 


63. 


+  2x2/  =  35  )  <-  2/ 

+  \^  xy  =  15 


fx  +  N/^^=15|  ^^^^      (a;  =  9. 

(.  J/  4-  Va-j/  =  10  )  (.  1/  =  4. 

(a;^  +  2/^=106|       ^^^      (a.  =  3,or3>/=T 


64.  Two  men  started  from  two  towns,  A,  o,nd  B,  and  tra- 
veltid  toward  each  other.  The  first  went  ^,  and  the  second, 
f  of  the  distance  between  the  two  towns,  when  the  men  were 
found  to  be  16  miles  apart;  required  the  distance  from  A  to  B 

Ans,  60  miles. 

65.  The  sum  of  two  numbers  is  80  ;  and  if  their  difference 
be  subtracted  from  the  less  and  added  to  the  greater,  the  re- 
sults will  be  as  1  to  7  ;  what  are  the  numbers  ? 

Ans.  80  and  50. 

66.  What  number  is  that  whose  fourth  part  exceeds  its  fifth 
part  by  ^j^^  ^  ^^^-  h 

67.  There  is  a  number  whose  3  digits  are  the  same ;  and  if 
from  the  number,  4  times  the  sum  of  the  digits  be  subtracted, 
the  remainder  will  be  297  ;  required  the  number. 

Ans.  333. 

68.  A  certain  number  increased  by  1,  is  to  the  same  number 
increased  by  4,  as  the  square  of  the  number  is  to  its  cube* 
what  is  the  number  ?  Ans,  2 


MISCELLANEOUS   EXAMPLES.  303 

69.  Fifty  gallons  of  wine  are  to  be  put  into  casks  of  two 
sizes ;  and  10  of  the  smaller  casks  and  2  of  the  larger,  or  5  of 
the  smaller  and  6  of  the  larger,  may  be  used ;  required  the 
capacity  of  a  cask  of  each  size. 

Ans.   Smaller,  4  gal.  ;  larger,  5  gal. 

70.  Two  men  have  the  same  income  ;  one  saves  one  tenth  of 
bis,  the  other  spends  $150  per  annum  more  than  the  first,  and 
at  the  end  of  five  years  finds  himself  $100  in  debt ;  what  is 
tht  income  of  each?  Aiis.  $loOO. 

71.  A  man  has  2  equal  flocks  of  sheep  ;  from  one  he  sells  a 
gheep^  and  from  the  other  b  sheep,  and  then  he  has  3  times  as 
many  remaining  in  the  latter  flock  as  in  the  former  ;  how  many 
did  each  flock  originally  contain  ?      A7is.  ^(Sa  —  b)  sheep. 

Note. — For  a  and  b,  take  uny  niimher  at  plensure,  such  that  Sa  —  h 
shall  be  divisible  by  2,  and  form  a  definite  problem.  For  instance,  as- 
sume a  =  12,  and  6  =  2,  then  the  number  in  each  flock  will  be  17. 

In  this  manner  numeral  problems  are  formed. 

72.  The  sum  of  two  numbers  is  72,  and  the  sum  of  their 
cube  roots  is  6  ;  what  are  the  numbers  ?  Ans.  64  and  8. 

73.  The  sum  of  the  squares  of  two  numbers,  multiplied  by 
the  sum  of  the  numbers  is  2336,  and  the  difference  of  their 
Bquares,  multiplied  by  the  difference  of  the  numbers  is  576  ; 
what  are  the  numbers  ?  Ans,  11  and  5. 

74.  The  product  of  two  numbers  multiplied  by  their  sum  is 
84  ;  and  the  sum  of  their  squares  multiplied  by  the  square  of 
their  product  is  3600  ;  what  are  the  numbers  ?     Aiis.  4  and  3. 

75.  A  market  man  bought  15  ducks  and  12  turkeys  for  105 
shillings,  and  he  obtained  2  more  ducks  for  IS  shillings,  thar. 
turkeys  for  20  shillings  ;  what  were  the  prices  ? 

Ans.   Ducks,  3  shillings ;  turkeys,  5  shillings. 

76.  The  sum  of  three  numbers  is  12  ;  one  third  of  the  sum 
of  the  first  and  second  is  equal  to  one  fifth  of  the  sum  of  the 
Becond  and  third  ;  and  the  second  minus  the  first  is  equal  to 
thf5  third  minus  the  second  ;   required  the  numbers  ? 

An&.  2,  4,  and  (I, 


gQ4  MISCELLANEOUS  EXAMPLES. 

77.  There  is  a  square  tract  of  land  containing  10  times  as 
many  acres  as  there  are  rods  in  the  fence  inclosing  it ;  how 
large  is  the  square  ?  Ans,  20  miles  square. 

78.  A  general  wishing  to  draw  up  his  regiment  into  a  square, 
found  by  trial  that  he  had  92  men  over  ;  he  then  increased 
each  side  by  2  men,  and  wanted  100  men  to  complete  the 
square  ;  how  many  soldiers  had  he  ?  Ans.  2301. 

79.  Some  bees  had  alighted  upon  a  tree  ;  at  one  flight  the 
square  root  of  half  the  number  went  away ;  at  another  |  of 
them  ;  and  two  bees  then  remained  ;  how  many  alighted  upon 
the  tree  ?  Ans.  72. 

80.  A  May-pole  is  56  feet  high.  At  what  distance  above 
the  ground  must  it  be  broken,  in  order  that  the  upper  part, 
clinging  to  the  stump,  may  touch  the  ground  12  feet  from  the 
foot?  Ans.  26 1  feet. 

81.  Divide  the  number  20  into  two  such  parts  that  the 
square  of  the  greater  diminished  by  twice  the  less,  shall  be 
equal  to  twice  the  square  of  the  less.  Ans.  12  and  8. 

82.  Three  numbers  are  in  arithmetical  progression  ;  their 
sum  is  27,  and  the  product  of  the  extremes  is  77  ;  required  the 
numbers.  Ans.  7,  9,  and  11. 

83.  A  gentleman  has  a  garden  10  rods  long  and  8  rods 
wide ;  he  would  lay  out  half  the  way  round  it  a  graveled  walk 
of  uniform  width  and  to  contain  ^  of  the  area  of  the  garden. 
How  wide  shall  the  walk  be  laid  out  ?      Ans,  13.4684- rd. 

84.  Two  numbers  are  in  the  proportion  of  a  to  6,  and  when 

c  is  added  to  each,  the  proportion  is  as  5  to  6  ;  what  are  the 

numbers  ?  ^  «^  -.        ho 

Ans.  -p-^ ;^,  and 


66  _  6a'  56  —  6a* 

85.  A  man  sold  a  horse  for  144  dollars,  and  gained  as  much 
per  cent  as  the  horse  cost  him.     What  did  the  horse  (^ost  him  ? 

Ans.  $80. 


MISCELLANEOUS  EXAMPLES.  305 

86.  Two  numbers  are  in  the  proportion  of  5  to  8,  and  if 
200  be  added  to  the  first,  and  120  to  the  second,  the  sums  will 
be  to  each  other  as  5  to  4 ;  what  are  the  numbers  ? 

Ans,  50  and  80. 

87.  A  person  bought  two  cubical  stacks  of  hay  for  £41,  each  of 
which  cost  as  many  shillings  per  cubic  yard,  as  there  were  yards 
in  a  side  of  the  other,  and  the  greater  stood  on  more  ground 
than  the  less  by  9  square  yards.  What  was  the  value  of  each 
stack  ?  Ans,  £16  and  £25. 

Let  X     =  a  side  of  the  greater  stack  in  yards ; 

And  y     =  a  side  of  the  other ; 

Then  x"    —y^  =  9  =  a-,  0) 

And  ar'y  +  xf  =  41-20  =  820  =  6         (2) 

From  (2)        x^    +  y^  =  —  (8) 

xy 

Squaring  (3)  x*    +  2j?y  +  2/*  =  ~r^  (« 

X  y 

Squaring  (i)  x^    —  2x'^y'^  -{-  y^  =  a^ 

Diff  4:xY  =~—  a\ 

X  y 

88.  A  farmer  had  3  more  cows  than  horses.  He  bought  2 
more  cows  and  sold  3  horses ;  and  he  then  had  5  times  as 
many  cows  as  horses.     How  many  had  he  at  first  ? 

Ans.  5  horses  and  8  cows. 

89.  Some  boys  on  a  frolic  incurred  a  bill  of  $12.  If  there  had 
been  two  more  in  the  company  each  would  have  been  charged 
30  cents  less.     How  many  were  in  company  ?  Ans.  8. 

90.  A  person  residing  on  the  bank  of  the  Ohio,  15  miles 
above  Cincinnati,  can  row  his  boat  to  the  city  in  2^  hours, 
but  it  requires  7i  hours  to  return.  With  what  force  can  he 
ruw  his  boat  in  still  water,  and  what  is  the  velocity  of  the  river  ? 

Alls.  Man  rows  4  miles  per  hour  ;  stream  flows  2  miles. 

2x  2x 

91.  What  is  the  value  of  a;  H ^—  divided  hj  x 

when  X  =  5i  ?  ^,^  ^ 

26*  u 


806  MISCELLANEOUS   EXAMPLES.      . 

.  92.  I  deposited  $200  in  a  savings  bank,  which  paid  6  per 
cent,  on  deposits,  interest  payable  semi-annually.  How  much 
did  my  money  amount  to  in  5  years,  the  interest  being  added 
to  the  principal  at  the  end  of  every  6  months  ? 

Note. — The  principal  is  the  first  term;  $1.03,  the  semi-annual  amount  of 
$1  at  the  rate  per  cent,  is  the  ratio ;  and  the  number  of  terms  minus  1  is  2 
times  the  number  of  years. 

Ans.  $268.'78  +  . 

93.  What  is  the  value  of  the  expression   -_ —  4- 

94.  Find  the  geometrical  mean  between  2a^h'^  and  24:a^b'^x'^ , 

12 8  — 

Ans.  4a^VSa  x  Vb^  x  Vx, 
05.  From  a  bag  of  money  which  contained  a  certain  sum,  was 
taken  $20  less  than  its  half;  from  the  remainder,  $30  less  than 
its  tliii'd  pai't ;  and  from  the  remainder,  $40  less  than  its  fourth 
part,  and  then  there  was  nothing  left.  What  sum  did  the  bag 
contain?  Ans.  1080. 

96.  Four  numbers  are  in  arithmetical  progression  ;  the  pro- 
duct of  the  first  and  third  is  27,  and  the  product  of  the  second 
and  fourth  is  72.     What  are  the  numbers  ? 

Ans.  3,  6,  9,  and  12. 

97.  A  merchant  gains  the  first  year,  15  per  cent,  on  his  capi- 
tal ;  the  second  year,  20  per  cent,  on  the  capital  at  the  close 
of  the  first ;  and  the  third  year,  25  per  cent,  on  the  capital  at 
the  close  of  the  second ;  when  he  finds  that  he  has  cleared 
$1000.50.     Required  his  capital.  A7is.  $1380. 

98.  The  sum  of  three  numbers  in  arithmetical  progression 
is  15,  and  their  product  is  80.     Required  the  numbers. 

Ans.  2,  5,  and  8. 

99.  Find  three  numbers  in  arithmetical  progression  such 
that  the  sum  of  their  squares  shall  be  2900,  and  the  product  of 
the  extremes  shall  be  less  than  the  square  of  the  mean  by 
100.  Ans.  20,  30,  and  40. 


MISCELLANEOUS  EXAMPLES.  307 

100.  What  number  is  that  which,  if  4  be  subtracted  from  it,  ^ 
of  the  remainder  will  be  7  ?  Ans.  25. 

101.  What  number  is  that  to  which  if  1  and  11  be  added  se- 
parately, the  sums  will  be  to  each  other  as  1  to  3  ? 

Ans.  4. 

102.  Given  - — --7-3  +  ~ — r-r^  ==  ^y  ^^  ^^^^  ^^^  ^^^^^  ^^'  ^• 

(a  +  bf        (a  +  by 

.  ah 

Ans.  X  ==  — ,-7. 
a  +  b 

103.  What  number  is  as  much  below  40,  as  three  times 
that  number  is  below  100  ?  Ans.  30. 

104.  Divide  400  into  two  such  parts,  that  the  sum  of  their 
square  roots  shall  be  28.  A^ns.  256  and  144. 

105.  A  man  sold  a  horse  for  a  dollars,  and  gained  as  much 
per  cent  as  the  horse  cost  him  ;  what  did  the  horse  cost  him  ? 

Ans.  (v^aHh~25J  10  —  50  dollars. 

106.  The  product  of  three  numbers  in  geometrical  progres- 
sion is  1728,  and  the  sum  of  the  first  and  third  is  40 ;  what 
are  the  numbers  ?  Ans.  4,  12,  and  36. 

107.  Two  quantities  are  to  each  other  as  m  to  n,  and  the 

difference  of  their  square  is  d^ ;  what  are  the  quantities  ? 

md  nd 

Ans.     /~~ :,     , -=r^ 

Vm?  —  n     vm-  —  1^ 

108.  The  sum  of  four  numbers  in  geometrical  progression 
is  85  ;  and  the  sum  of  the  first  two  is  to  the  sum  of  the  second 
two  as  1  to  16  ;  what  are  the  numbers  ? 

Ans.  1,  4,  16,  and  64. 

109.  Th.e  base  of  a  right  angled  triangle  is  20  rods,  and  the 
perpendicular  and  hypotenuse  are  to  each  other  as  5  to  7 ; 
what  is  the  length  of  the  perpendicular  and  what  the  area  of 
the  triangle  ? 

Ans.  Perpendicular,  20.4124  f  rods;  area,  204.124 -fs(^.  ^-ods. 


80.8  MISCELLANEOUS   EXAMPLi5S. 

110.  The  extremes  of  a  geometrical  series  are  2  and  3*744, 
and  the  ditference  of  the  first  and  second  term  is  to  the  difference 
of  the  third  and  fourth  as  1  to  9.  liequired  the  sum  of  the 
Beries.  -^^^s.  5615. 

111.  The  sum  of  two  numbers  added  to  the  sum  of  their 
squares  is  18,  and  10  times  their  product  is  60  ;  what  are  the 
numbers  ?  Ans.  2  and  3,  or  v^3  —  3,  and  —  Vo  —  3. 

112.  The  sum  of  two  numbers  is  to  their  difference  as  4  to 
1,  and  the  sum  of  their  cubes  is  152  ;  what  are  the  numbers  ? 

Ans.  3  and  5. 

113.  Insert  9  arithmetical  means  between  6  and  36. 

Ans.  9,  12,  15,  &c. 

114.  At  what  rate  per  cent,  will  a  dollars  gain  as  much  in 
4  years  at  simple  interest,  as  in  2  years  at  compound  interest  ? 

Ans.  200  percent. 

115.  How  many  terms  of  the  series,  .034,  .0344,  .0348,  &c.,  . 
will  amount  to  2.748?  Ans.  60. 

116.  How  much  will  $230  amount  to  in  12  years,  at  6  per 
cent.,  simple  interest  ? 

Note.  —  The  number  of  terms  will  evidently  be  1  greater  than  the 
number  of  years. 

Alls.  $395.60. 

117.  What  is  the  sum  of  n  terms  of  the  series,  3, 3^-,  ^,  &c.  ? 

Ans.  (n  +  17)^. 

118.  I  lent  a  certain  sum  at  7  per  cent.,  simple  interest,  and 
at  the  end  of  5  years  received,  in  principal  and  interest,  $317.79 ; 
what  was  the  sum  lent  ?  Ans.  $235.40. 

119.  If  6  be  the  first  term  of  a  geometrical  series,  and  4374 
the  7th  term,  what  is  the  ratio,  and  what  are  the  other  terms  ? 

Ans.  Ratio  3  ;  whence  18,  54,  &c.,  the  other  terms  of 
the  series. 


MISCELLANEOUS  EXAMPLES.  3()9 

120.  Sold  a  horse  for  $175.50,  taking  a  uote  drawing 
interest  at  6  per  cent.  Not  needing  the  money,  I  did  not 
collect  the  note  until  the  end  of  6  years ;  what  amount  did  I 
collect?  Ans.  $238.68. 

121.  The  sum  of  the  extremes  of  4  numbers  in  geometrical 
progression  is  35,  and  the  sum  ©f  the  means  is  30 ;  what  are 
the  numbers  ?  Ans.  8,  12,  18,  27. 

122.  I  own  a  mortgage  of  $875  on  a  farm,  due  in  6  years, 
at  6  per  cent,  interest,  payable  annually.  If  no  part  of  the  moit. 
gage  or  interest  is  paid  until  the  end  of  the  6  years,  how  muck 
will  be  the  auiountdue  at  compound  interest  ?  Ans.  $1241.20 -f. 

123  Three  times  the  product  of  two  numbers  is  equal  to  their 
difference  multiplied  by  the  difference  of  their  squares.  Also, 
45  times  the  square  of  the  product  is  equal  to  the  difference 
of  their  4th  powers  multiplied  by  the  difference  of  their  squares ; 
what  are  the  numbers  ?  Ans.  4  and  2. 

124.  The  principal  is  $300,  the  time  3  years,  and  the  rate  6 

per   cent.,   compound   interest,   semi-annually;    what  is  the 

amount  ? 

Note.  —  The  number  of  terms  is  1  more  than  2  times  the  number  of 
years;  and  the  ratio  is  1.03. 

Ans.  $358.2156. 

125.  The  length  of  a  plat  of  ground  is  4  rods  more  than  its 
breadth ;  and  the  number  of  square  rods  in  its  area  is  equal  to 
the  number  of  rods  in  its  perimeter.  Required  the  length  ard 
breadth.  (Length,  6.8284  +  rods. 

iBreadth,  2.8284  +  rods. 

126.  Find  the  side  of  a  cube  which  shall  contain  as  many 
solid  units  as  there  are  linear  units  in  the  distance  between  its 
two  opposite  corners.  j^^g^   ^/g^ 

127.  Find  two  numbers,  such  that  their  product  shall  be 
equal  to  4  times  their  difference ;  and  the  difference  of  their 
squares  shall  be  9  times  the  sum  of  the  numbers. 

Ans,  3  and  12. 


glO  MISCELLANEOUS  EXAMPLES. 

128.  A  and  B  together  carried  90  eggs  to  market,  and  sold  at 
different  prices,  each  receiving  the  same  sum.  Had  A  taken 
as  many  as  B  he  would  have  received  32  cents  for  them.  Had 
B  taken  as  many  as  A,  he  would  have  received  50  cents  for 
them  ;  how  many  did  each  take  to  market  ? 

Ans.  A  50,  B  40. 

129.  Find  two  numbers,  such  that  the  difference  of  tlicir 
scjuares  shall  be  12  ;  and  3  times  the  square  of  the  greater 
minus  twice  their  product  shall  be  32.  Ans.  4  and  2. 

130.  The  product  of  5  numbers  in  arithmetical  progression 
is  945,  and  the  sum  of  the  numbers  is  25  ;  what  are  the  num- 
bers? Ans.  1,  3,  5,  7,  9. 

13 1.  A  man  has  two  unequal  measures.  If  he  lay  out  a  plat 
of  ground  having  the  greater  measure  for  its  length,  and  the 
less  for  its  breadth,  it  will  contain  40  square  feet ;  but  if  he 
lay  out  a  plat  having  twice  the  greater  measure  and  once  the 
less  for  its  length,  and  once  the  greater  and  twice  the  less  for 
its  breadth,  it  will  contain  432  square  feet.  How  many  feet 
in  length  is  each  measure  ?  Ans.  Less,  4  ;  greater,  10. 

132.  A  carpenter  agreed  to  live  with  a  farmer  during  the 
winter,  on  the  condition  that  for  everyday  he  worked  he  should 
receive  $1.50,  and  for  every  day  he  was  idle  he  should  forfeit 
65  cents.  At  the  expiration  of  129  days  they  settled,  and  the 
carpenter  received  nothing  ;  how  many  days  did  he  work,  and 
how  many  was  he  idle  ? 

Ans.  He  worked  39  days,  and  wa«;  idle  90  days. 

133  Find  three  numbers  such  that  the  product  of  the  first 
and  second  shall  be  6 ;  of  the  first  and  third,  8  ;  and  of  the 
second  and  third,  12.  Ans.  2,  3,  and  4 

134.  In  a  plane  triangle  the  base  is  50  feet,  the  area  600  feet, 
and  the  difference  of  the  sides  10  feet ;  required  the  sides  and 
perpendicular.         Ans.  Sides,  ^0  and  40  ;  perpen.  24  feet. 


MISCELLANEOUS  EXAMPLES.  ^H 

135.  There  are  four  numbers  such  that  the  product  of  tho 

first,  second,  and  third  is  a ;  the  product  of  the  first,  second, 

and  fourth  is  b ;  the  product  of  the  first,  third,  and  fourth  is 

c;  and  the  product  of  the   second,  third,  and  fourth  is  d, 

Required  the  numbers. 

,         ^^  abed   -^^abcdi  -^ abed    \^abcd 

Ans.   — -r^— , ,  — -— , . 

d  0  b  a 

136.  A  is  a  traveler  11  miles  in  advance  of  B,  and  travels  4 
miles  per  hour ;  B  starts  to  overtake  him,  and  travels  4-^  miles 
the  first  hour,  4|  the  second,  and  5  the  third,  increasing  his 
rate  :^  of  a  mile  per  hour ;  how  many  hours  before  he  will 
overtake  A  ?  ^?7S.  8. 

137.  The  base  of  a  right  angled  triangle  is  6,  and  the  three 
Bides  are  in  arithmetical  j)rogression  ;  what  are  the  sides  ? 

Ans.  6,  8,  10,  or  4i  6,  7  J. 

138.  Two  squares  contain  together  an  area  of  52  inches ; 
and  twice  the  difference  of  their  area  is  equal  to  the  number 
of  inches  in  the  perimeters  of  the  two.  Required  the  con- 
tents, of  each.  .  (Greater,  86  inches; 

I  Less,       16  inches. 

139.  A  certain  number,  consisting  of  two  digits,  is  equal  to 
twice  the  product  of  its  digits ;  and  if  27  be  added  to  the 
number,  its  digits  will  be  inverted.     Required  the  number. 

Ans,  36, 

140.  The  sum  of  three  numbers  in  arithmetical  progression 
is  I,  and  the  sum  of  their  reciprocals  is  7^- ;  what  are  the 
numbers  ?  Ans.  ^,  \,  and  |. 

141.  Find  three  geometrical  means  between  -^  and  f. 

Ans.  ^>/6',  ^,  and  ^v^6. 

142.  There  are  three  quantities  related  as  follows  :  the  sum 
of  the  squares  of  the  first  and  second,  added  to  the  first  and 
second,  is  18  ;  the  sum  of  the  squares  of  the  first  and  third, 
added  to  the  first  and  third,  is  26 ;  and  the  sum  of  the  squares 


gio  MISCELLANEOUS   EXAMPLES. 

of  the  second  and  third,  added  to  the  second  and  third,  is  32 
Required  the  quantities. 

r  1st,  2,  or  —  3  ; 

Ans,  <  2d,  3,  or  —  4  ; 

I  3d,  4,  or  —  5. 

143.  The  compound  interest  of  a  certain  sum  of  money  for 
8  years  was  $864  ;  and  the  interest  for  the  first  year  was  to  the 
interest  which  accrued  the  third  year  as  25  to  36.  Required 
the  sum  at  interest.  Ayis.  $500. 

144.  Find  3  numbers  in  arithmetical  progression  such  that 
their  sum  shall  be  36,  and  4  added  to  the  product  of  the  ex- 
tremes shall  be  equal  to  the  square  of  the  mean. 

Ans,  10,  12,  and  14. 

145.  Find  three  numbers  in  geometrical  progression  whose 
sum  shall  be  52,  and  the  sum  of  the  extremes  to  the  square 
of  the  mean  as  10  to  36.  Ans,  4,  12,  and*  36. 

146.  There  are  three  numbers  in  geometrical  progression ; 
their  continued  product  is  1,  and  the  difference  of  the  first  and 
second  is  to  the  difference  of  the  second  and  third  as  1  to  3  ; 
what  are  the  numbers  ?  Ans.  ^,  1,  and  3. 

147.  There  are  three  numbers  in  geometrical  progression, 
such  that  three  times  the  first,  twice  the  second,  and  once  the 
third,  taken  in  order,  form  an  arithmetical  series  ;  and  also  the 
first,  the  second  increased  by  8,  and  the  third,  taken  in  order, 
form  an  anthnietical  series.     What  are  the  numbers  ? 

Am,  4, 12,  and  ?6. 


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